The idea is to try to use this theorem, but in essence it will be just a leading force into simplifying it all considerably.

*Theorem*

$e^x$ is an integer iff for every $n>ex$

$$1-\{\sum_{k=0}^{n}\{\frac{x^k}{k!}\}\} \leq \frac{2x^{n+1}}{(n+1)!}$$

This is easy to derive and 19th century mathematicians had tools to find that it is sufficient to take $ex$ terms in order to achieve the precision less than $1$ and the right part is a crude estimation of the error term.

However in order to make this more precise, we find an integer as close to $(\pi^\pi)^{\pi^\pi}$ as we can. Since we know $\pi^3 \approx 31$, the candidate is simply $31^{31}$ which is not that difficult to calculate after all:

$$31^{31}=\frac{(((((31)^2)^2)^2)^2)^2}{31}$$

Using that we are reducing the calculation from $300$ terms to about $80$ terms of working with smaller numbers. Yet we can do better:

$$\pi^{\pi+1}\ln(\pi)-31\ln(31) \approx 25$$

Going into this direction we can try to find max integer solution of

$$x^x < e^{24}$$
$$x = 10$$

That leaves

$$\pi^{\pi+1}\ln(\pi)-31\ln(31)-10\ln(10) \approx 1.64$$

making

$$(\pi^\pi)^{\pi^\pi}=31^{31}10^{10}e^{\pi^{\pi+1}\ln(\pi)-31\ln(31)-10\ln(10)}$$

Further

$$(\pi^\pi)^{\pi^\pi}=31^{31} 10^{10} 2^{2} e^{{\pi^{\pi+1}}\ln(\pi)-31\ln(31)-10\ln(10)-2\ln(2)}$$

And finally

$$1<e^{\pi^{\pi+1}\ln(\pi)-31\ln(31)-10\ln(10)-2\ln(2)} <2$$

As we are dealing with small numbers the idea is very obvious. Express

$$\pi^{\pi+1}\ln(\pi) = r+\sum_{k=1}^{m}a_k\ln(b_k), a_k \in \mathbb{N},b_k \in \mathbb{N}, 0 \leq r < \ln(2)$$

Once this is done it is only $\{e^r\}$ that decides if the expression is an integer or not which we can calculate to almost any desirable precision or limited but sufficient one manually. Still we need to do the final multiplication.

Notice that even without the final multiplication, just by calculating $\{e^r\}$ we may confirm beyond doubt that the number cannot be rational number (which is the only way of making the above expression an integer), or that it is not a correct rational number if it happens to be rational, for which we would not have to multiply with the large number $31^{31} 2^{2}$ at all. A nice thing to have is $10^{10}$ if we work in decimal notation, but any integer pair $p\ln(q)$ would do equally well.

Simplifications can be revolving around these two ideas, of course, starting from instantly trying to find the best matching $10^d$ to still using the series with a couple of terms, or trying out only small primes in order to get the simplest possible proof that $e^r$ is not rational and so on.

For example

$$\pi^{\pi+1}\ln(\pi) = 189\ln(2)+r$$

Now you calculate $e^r$ to about 60-digit precision. The worst case scenario is that it is of the form $\frac{p}{2^{189}}$ which would be obvious that it is not from its form. (Not that mathematicians were not capable of using binary form I doubt dealing constantly just with $0$ and $1$ would be beneficial. Still it is a very nice feature to have that $r$ has to stop in binary representation.)

Finally we pick

$e^{\pi^{\pi+1}\ln(\pi)- 56\ln(10)} = 8.874551721831242958746314552254346026884128667654661250051588548428...$

proving it is not an integer. The precision is about 65 digits and it does not end with the stream of $0$'s or possibly $9$'s.

19th century mathematicians were much smarter than this sample of ideas for sure.

Not to forget then about the "and now" part.

All we need is to prove that

$$e^{\pi^{\pi^\pi}\ln(\pi)-2213275961092431622\ln(2)}$$

is irrational. Piece of cake for 22nd century mathematician. :)

The most likely future proof is going to claim that

$$ \{ 1,^{n}\pi | n \in \mathbb{N} \} $$

are all linearly independent over $\mathbb{Q}$, where $^{n}x$ is tetration. Meaning none of exponents is an integer.