Shortest proof of \(e\)'s irrationality:

If \(e\) were rational, then
\[e=\dfrac{m}{n}=\displaystyle\sum_{k=0}^n\dfrac{1}{k!}+\sum_{k=n+1}^\infty\dfrac{1}{k!};\ (m,n)\in\mathbb{Z}_+^2\]

Multiplying both sides by \(n!\), we get
\[m(n-1)!-\displaystyle\sum_{k=0}^n\dfrac{n!}{k!}=\sum_{j=1}^\infty\dfrac{n!}{(n+j)!}<\sum_{j=1}^\infty\dfrac{1}{(n+1)^j}=\dfrac{1}{n}\]
The left side is an integer but \(0<\tfrac{1}{n}<1\), which contradicts the assumption. Hence, \(e\) is irrational. \(\blacksquare\)