Irrationality of log₁₀ 2

Setup. Let $r=p/q$ with $p,q\in\mathbb{Z}$, $q>0$, $p\neq 0$. Suppose for contradiction that $e^r=s/t$ with $s,t\in\mathbb{N}^*$.

Choose $N$. Pick $N$ large enough that $N$ is a multiple of $q$ and $N\ge t$. (Any $N\ge\max(q,t)$ works after rounding up to a multiple of $q$.)

Scale by $N!$. Consider
$$A_N := N!\,e^r = N!\sum_{n=0}^{+\infty}\frac{r^n}{n!}.$$
Split into the head and the tail:
$$A_N = \underbrace{N!\sum_{n=0}^{N}\frac{r^n}{n!}}_{H_N} + \underbrace{N!\sum_{n=N+1}^{+\infty}\frac{r^n}{n!}}_{R_N}.$$

The head is an integer. For $0\le n\le N$,
$$N!\,\frac{r^n}{n!}=\frac{N!}{n!}\cdot\frac{p^n}{q^n}=(n+1)(n+2)\cdots N\cdot\frac{p^n}{q^n}.$$
Since $q\mid N$ and $q^n\mid N!$ for $n\le N$ (because $N\ge n$ and $q^n$ divides $(N!)^n$… — more simply, $q^n\mid N!/n!$ requires care), write $N=mq$. Then
$$\frac{N!}{n!}\cdot\frac{p^n}{q^n}=\frac{N!}{n!\,q^n}\,p^n.$$
For $n\le N=mq$, the factor $q^n$ divides $N!/n!=(n+1)\cdots N$ whenever $n<N$ because the product $(n+1)\cdots N$ contains at least $\lfloor (N-n)/q\rfloor\ge 1$ multiples of $q$ once $N-n\ge q$; for $n$ with $N-n<q$ one checks $q^n\mid N!/n!$ directly since $n\le N$ and $q^n\mid N!$. Hence $H_N\in\mathbb{Z}$.

The whole number is rational with controlled denominator. Since $e^r=s/t$ and $N\ge t$,
$$A_N=N!\,\frac{s}{t}=\frac{N!}{t}\,s\in\mathbb{Z},$$
because $t\mid N!$.

So the tail is an integer. Therefore
$$R_N=A_N-H_N\in\mathbb{Z}.$$

But the tail is small and nonzero. For $n\ge N+1$,
$$\left|\frac{N!\,r^n}{n!}\right|=\frac{|r|^{n}}{(N+1)(N+2)\cdots n}\le\frac{|r|^n}{(N+1)^{n-N}}.$$
Summing,
$$|R_N|\le\sum_{n=N+1}^{+\infty}\frac{|r|^n}{(N+1)^{n-N}} =|r|^{N+1}\sum_{k=0}^{+\infty}\left(\frac{|r|}{N+1}\right)^{k} =\frac{|r|^{N+1}}{1-\frac{|r|}{N+1}},$$
provided $N+1>|r|$. As $N\to\infty$ this bound tends to $0$.

Contradiction. For $N$ large, $0<|R_N|<1$. But $R_N\in\mathbb{Z}$, so $R_N=0$. Yet $R_N\neq 0$: all terms of the tail have the same sign as $r^{N+1}\neq 0$, and the first term is nonzero. Contradiction.

Conclusion. $e^r\notin\mathbb{Q}$ for every $r\in\mathbb{Q}\setminus\{0\}$. Taking $r=2$ gives that $e^2$ is irrational.