$e=\lim\limits_{k\rightarrow\infty}\sum\limits_{n=0}^{k}\frac{1}{n!}$
Define f:
$f(k)=\sum\limits_{n=0}^{k}\frac{1}{n!}$
$e=\lim\limits_{k\rightarrow\infty}f(k)$
$f(k+1)=\sum\limits_{n=0}^{k+1}\frac{1}{n!}=f(k)+\frac{1}{(k+1)!}$
1/(k+1)! is rational so if f(k) is rational so is f(k+1).
f(0)=0 is rational so f(k) is rational for all k>0.
$\infty>0$
e is rational.

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$\lim\limits_{k\rightarrow\infty}(e-f(k))=\lim\limits_{k\rightarrow\infty}\sum\limits_{n=k+1}^{\infty}\frac{1}{n!}=\frac{1}{\infty!}=0$
in the limit the difference between e and f(k), which is rational, is zero.