Sorry this answer became so long. If you want the two-minute answer, just
read the **Terminology** then skip down to the **Answers to the Questions**.
You can then fill in the details as desired.

### Terminology

- Let $A, B, C, A_i$ denote the blue-eyed islanders.
- Let $A_i^*$ denote the proposition that $A_i$ has blue eyes (which does not imply that $A_i$ knows this).
- We'll use the standard symbols and rules of propositional logic.
- $A\leadsto P$ means islander $A$ knows that $P$ is the case, where $P$ is a proposition. The operator "$\leadsto$" is right-associative and has higher precedence than the logical implication operator "$\Rightarrow$".
- $\cal O$ is the proposition that there is at least one blue-eyed islander. This is what the Guru announces on day 1.

## 1 Blue-eyed Islander

To introduce use of this notation, let's briefly go over what happens when there is 1 blue-eyed islander $A$.

### Day 1

After the Guru makes her announcement, it is the case that
$$A\leadsto{\cal O}.\tag{1.1.1}$$
Since $A$ sees no islanders with blue eyes, she concludes it's her:
$$A\leadsto A^*.\tag{1.1.2}$$
This fact entitles $A$ to leave the island on day 1, so we are done
with the case of the 1 Blue-eyed Islander.

## 2 Blue-eyed Islanders

Before the Guru makes her announcement, the following statements
can be said about blue-eyed islanders $A$ and $B$:
$$A\leadsto{\cal O}\\
B\leadsto{\cal O}.\tag{2.0.1}$$
This follows simply from the fact that they can each see one other
blue-eyed person on the island.

### Day 1

After the Guru announces $\cal O$, it is the case that
$$A\leadsto B\leadsto{\cal O}\tag{2.1.1}$$
as well as the one other permutation of this: $B\leadsto A\leadsto\cal
O$. The above reads "$A$ knows that $B$ knows that $\cal O$." It
is important to grasp at this point that this fact wasn't the case
prior to the Guru's announcement. Even though both $A$ and $B$ knew
there was at least one blue-eyed person on the island, $A$ didn't
know that $B$ knew this, because for all $A$ knows, she may have
brown eyes.

(2.1.1) can be written as
$$A\leadsto B\leadsto(A^*\vee B^*).\tag{2.1.2}$$
The substitution ${\cal O}\mapsto A^*\vee B^*$ is valid in this
particular context, because everyone else on the island other than
$A$ and $B$ do not have blue eyes, which is known to both $A$ and
$B$, and $A$ knows that $B$ knows this.

(2.1.2) can be written as
$$A\leadsto(\neg A^*\Rightarrow B\leadsto B^*)\tag{2.1.3}$$
which will be useful for Day 2. To prove this, the following axiom
is needed:

**Knowledge Conjunction Axiom**
$$((A_i\leadsto P)\;\wedge\;(A_i\leadsto Q))\;\;\Leftrightarrow\;\;
(A_i\leadsto(P\wedge Q))$$
$A_i$ knows $P$ and $A_i$ knows $Q$ if and only if $A_i$ knows $P$ and $Q$.

The proof for (2.1.3) is of primary importance, as it readily
generalizes to any number of blue-eyed islanders and days,
so a detailed proof is given here for the interested reader.

- $A\leadsto(\neg A^*\Rightarrow(B\leadsto\neg A^*))$
- $A\leadsto((B\leadsto\neg A^*)\Rightarrow(B\leadsto\neg A^*))$
- $A\leadsto(((B\leadsto\neg A^*)\Rightarrow(B\leadsto\neg A^*))\wedge
(B\leadsto(A^*\vee B^*)))$
- $A\leadsto((B\leadsto\neg A^*)\Rightarrow((B\leadsto\neg A^*)\wedge
(B\leadsto(A^*\vee B^*))))$
- $A\leadsto((B\leadsto\neg A^*)\Rightarrow(B\leadsto(
\neg A^*\wedge(A^*\vee B^*))))$
- $A\leadsto((B\leadsto\neg A^*)\Rightarrow(B\leadsto B^*))$
- $A\leadsto(\neg A^*\Rightarrow B\leadsto B^*)$

Step-by-step justifications:

- $A$ knows that if she doesn't have blue eyes, then $B$ will know this.
- $P\Rightarrow P$ tautology.
- Knowledge Conjunction Axiom of step 2 with (2.1.2).
- $((P\Rightarrow P)\wedge Q)\;\Rightarrow\;(P\Rightarrow(P\wedge Q))$
tautology applied to step 3.
- Knowledge Conjunction Axiom applied to step 4.
- Disjunctive syllogism applied to step 5.
- Knowledge Conjunction Axiom applied to steps 1 and 6, and transitivity
of $\Rightarrow$.

This just delineates in detail what many people can reason without
the symbolic logic, which is the fact that $A$ knows that if she
doesn't have blue eyes, then $B$ will know he does. The value of
this formalism is that it extends readily into more complicated
scenarios where our intuition may have trouble keeping up.

### Day 2

No one left the island on Day 1, so no one knew they had blue eyes.
In particular,
$$\neg(B\leadsto B^*)\tag{2.2.0}$$
otherwise $B$ would have left. This fact is publicly known, so in
particular $A$ knows it:
$$A\leadsto\neg(B\leadsto B^*)\tag{2.2.1}.$$
Combining this with (2.1.3) via the Knowledge Conjunction Axiom gives
$A\leadsto((\neg A^*\Rightarrow B\leadsto B^*)\wedge\neg(B\leadsto B^*)).$
Modus tollens yields
$$A\leadsto A^*.\tag{2.2.2}$$
This is $A$'s ticket off the island, so she leaves today. These
arguments for both days are symmetric in $A$ and $B$, so apply to
$B$ as well. Both blue-eyed islanders leave the island on Day 2.

## 3 Blue-eyed Islanders

Before the Guru makes her announcement, $C\leadsto\cal O$,
$B\leadsto\cal O$, and $A\leadsto\cal O$. In addition,
$$A\leadsto B\leadsto{\cal O}\\
B\leadsto C\leadsto{\cal O}\\
C\leadsto A\leadsto{\cal O}.\tag{3.0.1}$$
For example, $A$ knows $B$ knows $\cal O$, because $A$ knows $B$
knows $C^*$.

### Day 1

After the Guru announces $\cal O$, it is the case that
$$A\leadsto B\leadsto C\leadsto{\cal O}\tag{3.1.1}$$
as well as the $3!-1=5$ other permutations of this in $(A,B,C)$.
It is important to pause at this point and understand that this was
not true prior to the Guru's announcement. Especially if one wishes
to understand what *quantified information* the Guru is actually
providing that each person didn't already have, this is it. Even
though everyone knew that everyone else knew $\cal O$, that's only
2 levels deep. It required the Guru's public announcement to get
to the 3rd level. $A$ did not know that $B\leadsto C\leadsto\cal
O$ prior to her announcement.

Though the symbols provide the formalism, one informal but ituitive
notion to consider is that of each person's "world"—the
information available to a person. The world as seen through the
eyes of $A$ is one in which there are 2 other blue-eyed people $B$
and $C$. Now consider the world of $B$ as considered by $A$. In
this world, there is only 1 person whom with certainty has blue
eyes: $C$. Moreover $C$ does not know if anyone else has blue eyes;
$C$ does not know $\cal O$ until the Guru speaks, whose announcement
penetrates through these worlds so that even in this doubly layered
consideration, $C\leadsto\cal O$. In other words, $A\leadsto B\leadsto
C\leadsto{\cal O}$. (Why am I reminded of the movie *Inception*?)

Moving forward as before, from (3.1.1),
$$A\leadsto B\leadsto C\leadsto(A^*\vee B^*\vee C^*)\tag{3.1.2}$$
using the substitution ${\cal O}\mapsto A^*\vee B^*\vee C^*$. This
is just representing the fact that any of $A$, $B$, or $C$ might
be the one the Guru was talking about, which everyone knows that
everyone knows etc. to arbitrary depth.

It follows from (3.1.2) that
$$A\leadsto B\leadsto(\neg(A^*\wedge B^*)\Rightarrow C\leadsto C^*).
\tag{3.1.3}$$
The proof for this takes on an analogous structure as the proof for
(2.1.3) above.

### Day 2

As before, no one left on Day 1, so it is concluded that
$\neg(C\leadsto C^*)$ (as well as for $A$ and $B$). Since this
is just as public as the Guru's announcement, everyone knows everyone
knows etc. this to arbitrary depth. In particular:
$$A\leadsto B\leadsto\neg(C\leadsto C^*)\tag{3.2.1}.$$
This fact, and its 5 other permutations, were not true until it was
publicly observed that no one left on the ferry the prior midnight.
Combining this with (3.1.3) via the Knowledge Conjunction Axiom,
and applying modus tollens as before, yields:
$$A\leadsto B\leadsto(A^*\vee B^*).\tag{3.2.2}$$
Now we are beginning to see a pattern here. This can be written as
$$A\leadsto(\neg A^*\Rightarrow B\leadsto B^*)\tag{3.2.3}$$
which is identical to (2.1.3).

One might wonder, since $A$ knows that $\neg(B\leadsto B)$, because
$B$ did not leave last night, can it be deduced from (3.2.3) that
$A\leadsto A$?

The answer is no, but to see this it must be noted when certain knowledge
was obtained. It would be more precise to write (3.2.3) as
$$A\leadsto_2(\neg A^*\Rightarrow B\leadsto_2 B^*)\tag{3.2.4}$$
where $\leadsto_k$ denotes knowledge on Day $k$. The knowledge described
in (3.2.3) was not known until Day 2, after observing that $C$ did not
board the ferry, which means $\neg(C\leadsto_1 C^*)$, or that $C$ did not
know she had blue eyes *on Day 1*. Similarly,
$$A\leadsto_2\neg(B\leadsto_1 B^*).\tag{3.2.5}$$
Thus modus tollens cannot be applied to (3.2.4) and (3.2.5) because
$B\leadsto_1 B$ and $B\leadsto_2 B$ are two different propositions.

### Day 3

No one left again the previous night, so
$$A\leadsto\neg(B\leadsto B^*).\tag{3.3.1}$$
Combined with (3.2.3), $A\leadsto A^*$ so $A$ can now leave the
island. Same reasoning applies to $B$ and $C$, so all 3 islanders
leave on Day 3.

## $n$ Blue-eyed Islanders

Assume $2<n$. Prior to the Guru's announcement, it is a fact that
$$A_1\leadsto A_2\leadsto\cdots
\leadsto A_{n-2}\leadsto A_{n-1}\leadsto{\cal O}\tag{4.0.1}$$
including all other combinations and permutations of this chain of
equal or lesser length, out of the the $n$ blue-eyed islanders $A_i$
for $i\in\{1,2,...,n\}$. Note that (4.0.1) includes only $n-1$
islanders, not $n$. This is because $A_1$ can imagine the world
through $A_2$'s eyes, who looks through $A_3$'s eyes, ..., who looks
through $A_{n-2}$'s eyes, who looks through $A_{n-1}$'s eyes, who
gazes upon $A_n$ but in this world no information is available to
guarantee any other blue-eyed person is on the island. So it cannot
be concluded in this $(n-1)$-nested world that $A_n\leadsto\cal O$.
Without the Guru's announcement, the longest chain of distinct
blue-eyed islanders that can be stated is one which includes no
more than $n-1$ islanders, such as (4.0.1).

### Day 1

Once the Guru announces $\cal O$, the chain can now include all $n$
blue-eyed islanders. The guru's statement is equivalent to $n!$
statements, which are all the permutations of $A_i$ in
$$A_1\leadsto A_2\leadsto\cdots
\leadsto A_{n-1}\leadsto A_n\leadsto{\cal O}.\tag{4.1.1}$$
In anyone's world, no matter how deep the levels, the knowledge of
$\cal O$ is always available.

Substituting as before for $\cal O$, this becomes
$$A_1\leadsto A_2\leadsto\cdots
\leadsto A_{n-1}\leadsto A_n\leadsto\bigvee_{i=1}^n A_i^*.\tag{4.1.2}$$
Using analogous steps to prove (2.1.3) from (2.1.2), it follows from
(4.1.2) that
$$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-2}\leadsto A_{n-1}\leadsto
\left(\neg\bigvee_{i=1}^{n-1}A_i^*\Rightarrow A_n\leadsto A_n^*\right)
.\tag{4.1.3}$$

### Day 2

As in previous scenarios, since $A_n$ in particular didn't leave,
it is publicly known that $\neg(A_n\leadsto A_n^*)$. Everyone
already knew this so what new information is there? The new
information may be expressed as another set of chain statements, of
all $n!$ permutations in $A_i$ of
$$A_1\leadsto A_2\leadsto\cdots\leadsto
A_{n-2}\leadsto A_{n-1}\leadsto\neg(A_n\leadsto A_n^*).\tag{4.2.1}$$
This wasn't the case until the previous ferry left with no passengers.

Combining (4.1.3) and (4.2.1) together, using the Knowledge Conjunction
Axiom and modus tollens yields
$$A_1\leadsto A_2\leadsto\cdots \leadsto A_{n-2}\leadsto A_{n-1}\leadsto
\bigvee_{i=1}^{n-1} A_i^*.\tag{4.2.2}$$
Following the same pattern as before, it can be deduced that
$$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-3}\leadsto A_{n-2}\leadsto
\left(\neg\bigvee_{i=1}^{n-2}A_i^*\Rightarrow A_{n-1}\leadsto A_{n-1}^*\right)
.\tag{4.2.3}$$

### Day $k$

Assume $1<k\le n$. On the previous night $A_{n-k+2}$ didn't leave,
so it is publicly known that $\neg(A_{n-k+2}\leadsto A_{n-k+2}^*)$.
The new information that wasn't previously available allows for all
permutations and combinations in $A_i$ of identical length of the
following statement to be made:
$$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-k}\leadsto
A_{n-k+1}\leadsto\neg(A_{n-k+2}\leadsto A_{n-k+2}^*).\tag{4.3.1}$$
Combined with the conclusions from the previous day, it is the case that
$$A_1\leadsto A_2\leadsto\cdots\leadsto A_{n-k}\leadsto
A_{n-k+1}\leadsto\bigvee_{i=1}^{n-k+1} A_i^*.\tag{4.3.2}$$
This can be proven from induction on $k$, which is omitted for
brevity but is of the same form as the proof for (2.1.3). Also if
one follows how (4.2.3) was derived from (4.2.2), (4.2.1) and (4.1.3)
then this will also outline how one can prove this via induction.

### Day $n$

(4.3.2) shrinks by one islander on each passing day, until finally
when $k=n$ we are left with
$$A_1\leadsto A_1^*.\tag{4.4.0}$$
Since these arguments have been symmetrical in all the $A_i$,
$$\forall i\in\{1,2,...,n\}:A_i\leadsto A_i^*.\tag{4.4.1}$$
On Day $n$, all blue-eyed islanders leave the island.

## Answers to the Questions

1) What is the quantified piece of information that the Guru provides that each person did not already have?

All $100!\approx 9.3\times 10^{157}$ permutations in $A_i$ of the statement
$$A_1\leadsto A_2\leadsto\cdots
\leadsto A_{99}\leadsto A_{100}\leadsto{\cal O}.\tag{5.1.1}$$
Everyone already knows ${\cal O}=\bigvee_{i=1}^{100}A_i^*$. That
is not the value of the Guru's announcement. It is that everyone
knows that everyone know that everyone knows ... that $\cal O$ is
the case, that is the new information provided by the Guru's
announcement that wasn't previously known. In contrast, if the Guru
were to tell all the islanders in private the same fact $\cal O$,
no islanders would be able to leave the island. So it is not simply
the information content of her words that we must look at; there
is additional information in knowing that everyone else heard her
too. Correspondingly, each day that passes provides a new piece
of information that is comparably subtle. When an islander doesn't
leave, it is like another public announcement, which includes more
information than just the fact that $A_i$ didn't leave last night.
It is the knowledge that everyone else knows too. This knowledge is
quantified in the answer to question 3 below. Eventually, after
100 days, these additional pieces of information will shrink (5.1.1)
down to a fact that the islander can act upon. Namely,
$A_i\leadsto A_i^*$ for all $i\in\{1,2,...,100\}$.

2) Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?

Yes, everyone knows that there are no less than 99 blue-eyed people
on the island. The key concept to this problem is the recursive
nature in which islanders deduce what they can, knowing only the
information that they bring with them as they consider the world
through one anothers' eyes. As $A_1$ does this from the perspective
of $A_2$ who sees through the eyes of $A_3$, ..., who sees through
the eyes of $A_{98}$, $A_{98}$ is left only to gaze upon and consider
what $A_{99}$ and $A_{100}$ can possibly know within a world of
such limited information. In this nested world 98 levels deep, we
cannot take for granted that islanders $A_1$ through $A_{98}$ have
blue eyes, just as we cannot take for granted that $A_1$ has blue
eyes when we consider only her point of view on all the rest of the
99 blue-eyed islanders. Therefore considering the logic of a 2
blue-eyed-inhabited island is a worthwhile consideration. When 2
days go by in which $A_{99}$ and $A_{100}$ don't leave the island,
then $A_1$ knows $A_2$ knows ... knows $A_{97}$ knows $A_{98}$ knows
that someone else other than $A_{99}$ and $A_{100}$ have blue eyes.
Now there's only 98 days to go.

If this escapes the intuition, then consider the case of 1, 2, and 3
blue-eyed islanders, and allow the logical formalism as deliniated
above to provide the scaffolding that extends the intuition.

3) Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?

Because they're not. Every day that passes, new information is
provided that wasn't previously known. It's not as simple as a
statement that islander $A_k$ didn't leave, because yes that was
already known and anticipated. There is additional information in
the knowledge that everyone else knows that everyone else knows
etc. that $A_i$ did not leave the island. To be precise, on day
$k$, for $k>1$, the new facts that weren't previously the case are
all $100!/(k-2)!$ permutations and combinations in $A_i$ of
$$A_1\leadsto A_2\leadsto\cdots\leadsto A_{100-k}\leadsto
A_{101-k}\leadsto\neg(A_{102-k}\leadsto A_{102-k}^*).\tag{5.1.2}$$
With each passing day $k$, it is these facts that whittle away at
the chain of knowledge (5.1.1) setup by the Guru for each islander
on Day 1.