e is irrational

Formula. For every $n$ there is $c\in(0,1)$ with
$$e=\sum_{k=0}^n\frac1{k!}+\frac{e^c}{(n+1)!}.$$

Assume $e=\frac pq$. Take $n\ge\max(q,3)$. Multiply by $n!$:
$$n!\,e=\sum_{k=0}^n\frac{n!}{k!}+\frac{e^c}{n+1}.$$
- The left side is an integer, because $q$ divides $n!$.
- $\sum_{k=0}^n\frac{n!}{k!}$ is an integer.

So $\dfrac{e^c}{n+1}$ is an integer. But $0<e^c<e<3$ and $n+1\ge4$, so
$$0<\frac{e^c}{n+1}<\frac34<1.$$
There is no integer strictly between $0$ and $1$: contradiction. Hence $e\notin\mathbb{Q}$.

Remark. This is Fourier's proof (1815). The same method does not work for $\pi$, which is also irrational, but whose proof needs more work.