\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = \lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\frac{1}{i!}\right) since:
(1) Since, it can be shown that both \lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n and \lim\limits_{n \to \infty}\left(\frac{1}{n!}\right), the only remaining point is to show that:
\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n =\lim\limits_{n \to \infty}1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \dots + + \frac{1}{n!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\dots\left(1 - \frac{n-1}{n}\right) = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!}