In my classes I sometimes have a contest concerning who can write the largest number in ten symbols. It almost never comes up, but I'm torn between two "best" answers: a stack of ten 9's (exponents) or a 9 followed by nine factorial symbols. Both are undoubtedly huge, but I haven't been able to produce an argument that one is larger (surely they aren't equal). Any insight into which of these two numbers is bigger would be greatly appreciated.

You can always define new symbols which give you higher number than before. You want perhaps to limit your alphabet to digits, addition, multiplication, exponentiation, factorial. – Asaf Karagila Feb 22 '11 at 15:20

It seems $9!!!!!!!!!$ is bigger. If you input both into wolfram alpha, then the factorial computation times out, while the $9^{999999999}$ yields a number with $954 242 509$ digits. This is not perfect since the way factorials are calculated may make this computation time out earlier, however. – Justin Feb 22 '11 at 15:29

3Try taking the logarithm several times, estimating the results using Stirling's formula: http://en.wikipedia.org/wiki/Stirling's_approximation – Qiaochu Yuan Feb 22 '11 at 15:30

9@Fdart17: $9^{999999999}$ is enormously smaller than the exponential tower of ten 9s, which is the number being considered here. – Chris Eagle Feb 22 '11 at 15:35

4Depending on what you're willing to allow without bracketing, neither of these is the best you can do with these symbols. $9!!!!!!!!!$ is smaller than $9^9!!!!!!!!$, while (tower of ten 9s) is smaller than (tower of nine 9s)!. – Chris Eagle Feb 22 '11 at 15:38

The mathematicians version of "Mine's bigger than yours"! – AnnanFay Feb 22 '11 at 18:09

5"(surely they aren't equal)": One way to see that the exponential tower is not equal to 9!!!!!!!!! is to note that the first is odd and the second is even. – Jonas Meyer Feb 22 '11 at 18:58

Is $9!!!!!!!!$ even standard notation? Aren't brackets necessary there? For example, $n!!\ne (n!)!$. That wouldn't allow you to use so many factorials. (Although your question still stands open but I'm clueless.) – Lazar Ljubenović Apr 05 '13 at 06:45
4 Answers
$n!$ grows more rapidly than $9^n$ (it grows approximately with $n^n$), so eventually it wins out; in fact, a few moments with Wolfram Alpha suggests that for all $n\gt 21$, $n! \gt 9^n$. This means that the best solution is mixed: after the first exponentiation (since $9^9$ is greater than $9!$), you're better off using factorials, giving the answer $9^9!!!!!!!!$. (Also, note that we're all assuming that the correct parentheses here are implicit, since there's a big difference between $\left(9^9\right)^9$ and $9^{\left(9^9\right)}$.)
On the other hand, if you're allowed to use moreorless stock mathematical notation, you may want to have a look at Knuth's arrow notation; $9\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow9$ is bigger by far than the rest of these. (And numbers even larger than this have come up in mathematics  have a look, for instance, at Graham's Number, which most conveniently uses the arrow notation in its definition.)
EDIT: In fact it's pretty straightforward to prove that the answer in the first paragraph, $9^9!!!!!!!!$, is the best that can be done (with these operations). First, as long as $a$ is at least two symbols (for this group of symbols), then $a^9 \lt 9^a\lt a!$, so adding a factorial symbol is always going to be better than adding a $9$ onto your tower. But what about structures like $9!^{9!}$? Well, if both $a$ and $b$ are at least two symbols long, then $a^b\lt \mathrm{max}(a,b)!!$; assuming for the moment that $b$ is larger (because if $a$ were larger, then $b^a\gt a^b$), then $b! \approx b^b$, and while this isn't enough to ensure that it's greater than $a^b$ (consider the case where $a$ = $b$ = $9!$), it's close enough that we can be certain taking a second factorial will be larger  so those extra symbols you spent on exponentiation should retroactively have just gone into more exclamation points, and this is enough to guarantee that $9^9!!!!!!!!$ is the largest possible combination of these particular operations.
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2What do you mean by "so eventually it wins out" ? Of course I agree that $9^9!!!!...$ is best possible, but what about the original structures the OP is asking about? I think you can you prove that for any number of characters $n$, the power tower of nines will be larger than the iterated factorials. – Eric Naslund Feb 22 '11 at 17:38

1+1 for mentioning Grahams Number. I always have little chuckle when I see it's definition. – Eric Naslund Feb 22 '11 at 17:50

1Sorry  by 'eventually it wins out' I just meant that for some n, n! will be greater than 9^n  but as Jonas points out, that doesn't imply that the iteratedfactorial will be larger than the iteratedexponential with one additional iteration. – Steven Stadnicki Feb 22 '11 at 18:14
Wolfram Alpha shows $$9^{9^9}\approx 10^{10^{8.56}}$$ while $$(9!)! \approx 10^{10^{6.26}}$$ Adding another character adds another 10 to the bottom of the stack without changing the upper exponent much at all. It will even do the full stack of 10. The exponents have 8.5678 atop the tower of 10's, while the factorials have 6.26949. So the exponents win.
Added: In Douglas Hofstadter's May, 1982 column "On Number Numbness" he declares these essentially equal. For numbers of this size, the first thing you should look at is how many times you have to take a log to make it reasonable, which is 9 for both of them. Then look at the number on top of the stack, which is the only one that matters. So it is like 9.85678 compared with 9.626949, which are very close.
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8+1 for being the only one to answer the OP's original question, namely that the tower of nines is larger than the factorials. In fact, I think you can you prove it is true no matter how many times we iterate. That is for any number of characters, say $n$, the power tower of nines will be larger than the iterated factorials. (I find that shocking since factorials is basically $n^n$) – Eric Naslund Feb 22 '11 at 17:32

1I have to admit, I find this really surprising too, but no less impressive for it! – Steven Stadnicki Feb 22 '11 at 18:26

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@Eric: thanks for noticing. Sometimes these questions take on a life of their own, but there was an original question that prompted it. – Ross Millikan Feb 23 '11 at 03:25

$9!!!$ is roughly $(10^{10^{6.26}})^{10^{10^{6.26}}}$, much greater than $9^{10^{10^{8.56}}}$ (most likely), so adding a character (a factorial in one case and an exponent to the tower in the other), seems to favor factorials here by quite a bit. – Mitch May 14 '11 at 17:51

@Mitch: but adding a new number to the bottom of the stack is quite effective. As I said, Alpha shows the stack of exponentials wins over factorials for the same number of characters all the way up. But repeated factorials of 9^9 are better yet – Ross Millikan May 14 '11 at 23:50

Just to clarify: the method of comparing numbers in the last paragraph does not originate with Hofstadter. It was discussed by Littlewood in his Miscellanies, and probably goes back even further historically. – Willie Wong Sep 05 '11 at 16:01
A thorough study of this question is given in the article "Exponential vs. Factorial" by Velleman (American Mathematical Monthly Vol. 113, No. 8, Oct. 2006, pp. 689704). In the motivating example there are 5 characters instead of 10. Of course, the conclusion is the same as in Steven Stadnicki's answer, that $9^9!!!!!!!!$ (or $9^9!!!$ in the case of 5 characters) is the largest possible.
The article also addresses your original question, proving that the exponential tower of $n$ $9$s is always larger than a $9$ with $n1$ factorials applied (and many more general statements).
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@Eric: Thanks. I just knew where to look because I remembered skimming the article when that Monthly was new. – Jonas Meyer Feb 22 '11 at 18:04
I find it quite sad that none of the answers has actually formally shown that
$$9^{9^{9^{9^{9^{9^{9^{9^{9^9}}}}}}}} > 9!!!!!!!!!$$
The proces is a bit long, but it it worth it in my opinion. First, a simple calculation shows $9! < 9^8$. Then $9!! < (9^8)^{9^8} = 9^{8\cdot9^8}$. Then $$9!!!< \left(9^{8\cdot9^8}\right)^{9^{8\cdot9^8}} < \left(9^{9^9}\right)^{9^{8\cdot9^8}}=9^{9^{8\cdot9^8}\cdot9^9}=9^{9^{8\cdot9^8+9}}$$
$$9!!!! < \left(9^{9^{9^9}}\right)^{9^{9^{8\cdot9^8+9}}}=9^{9^{9^{8\cdot9^8+9}}\cdot 9^{9^9}}=9^{9^{9^{8\cdot9^8+9}+9^9}}<9^{9^{9^{8\cdot9^8+10}}}$$
Now, some notation. Let us use $f(x)$ for $9^x$ and $f^k(x)=f(f(... f(f(x))...))$. Let $x!^k$ denote $x!!!...!!!$ with $k$ factorials.
Clearly, $f(x)+1<f(x+1)$. By induction, $f^t(x)+1<f^t(x+1)$ for any $t$.
Now we proceed by induction to show that $9!^k < f^{k1}(8\cdot9^8+k+6)$. The base case has been given above. Assume $4 \leq k \leq 9^87$.
$$9!^{k+1} < \left(f^{k1}(8\cdot9^8+k+6)\right)^{f^{k1}(8\cdot9^8+k+6)}= 9^{f^{k2}(8\cdot9^8+k+6)f^{k1}(8\cdot9^8+k+6)}<9^{9^{f^{k3}(9^9)+f^{k2}(8\cdot9^8+k+6)}}<9^{9^{9\cdot f^{k2}(8\cdot9^8+k+6)}}<9^{9^{9^{f^{k3}(8\cdot9^8+k+6)+1}}}<9^{9^{9^{f^{k3}(8\cdot9^8+k+7)}}}=f^k(8\cdot9^8+k+7)$$
This completes the induction hypothesis. Hence we have shown the asked.
Now for some fun: The largest number using ten symbols. We can also use shortened Knuth's up arrow notation, $9 \uparrow^{9 \uparrow^{9 \uparrow^{9} 9} 9} 9$ has ten symbols.
Conway chained arrow notation, $9 \rightarrow 9 \rightarrow 9 \rightarrow 9 \rightarrow 9^9$.
Rado's sigma function, $\Sigma(\Sigma(\Sigma(9)))$.
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