I find some proof here, but can I prove it in the following way?
Assume $e$ is rational, then $e=\frac{p}q$, both $p$ and $q$ are positive integers. By Lagrange's Remainder Theorem,
$$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+\frac{1}{q!}x^q+\frac{e^c}{(q+1)!}x^{q+1}, ~~~~~\text{where} ~~~|c|<|x|$$
Let $x=1$ and multiply both sides with $q!$
$$e\cdot q!=q!+q!+q!\frac{1}{2!}+q!\frac{1}{3!}+...+1+\frac{e^c}{q+1}, ~~~~~\text{where} ~~~|c|<1$$
So we have
$$\text{some integer}=\text{some integer} + \frac{e^c}{q+1}$$
Since $0<e^c<3$ and $q+1>3$, (easy to exclude $q\neq 1, q\neq 2$)
So the tail $0<\frac{e^c}{q+1}<1$, and we get a contradiction.