How to show that $ \sum_{n = 0}^{\infty} \frac {1}{n!} = e$? 
How to show that $\sum\limits_{n = 0}^{\infty} \frac {1}{n!} = e$ where $e = \lim\limits_{n\to\infty} \left({1 + \frac 1 n}\right)^n$?

I'm guessing  this can be done using the Squeeze Theorem by applying the AM-GM inequality. But I can only get the lower bound. 
If $S_n$ is the $n$th partial sum of our series then, 
$$ \left({1 + \frac 1 n}\right)^n = 1 + n\cdot\frac{1}{n} + \frac{n(n - 1)}{2}\cdot \frac{1}{n^2} + \cdots + \frac{n(n - 1)\ldots(n - (n -1))}{n!} \cdot \frac{1}{n^n} $$
$$ \le 1 + \frac{1}{2!} + \cdots + \frac{1}{n!} =S_n $$
How can I show that $S_n \le $ a subsequence of $\left({1 + \frac 1 n}\right)^n$ or any sequence that converges to $e$?

Let $a_n = \left(1+ 1/n\right)^n$.  By the binomial theorem,
$$a_n = 1 + 1 + \frac1{2!}\left(1- \frac1{n}\right)+ \ldots +\frac1{n!}\left(1- \frac1{n}\right)\ldots\left(1- \frac{n-1}{n}\right)\leq \sum_{k=0}^{\infty}\frac1{k!}=e,$$
The sequence $a_n$ is increasing and bounded, so it converges.
Hence, $$\lim_{n\rightarrow \infty} a_n \leq e.$$
For $m < n,$
$$a_n \geq 1 + 1 + \frac1{2!}\left(1- \frac1{n}\right)+ \ldots +\frac1{m!}\left(1- \frac1{n}\right)\ldots\left(1- \frac{m-1}{n}\right).$$
Take the limit as $n \rightarrow \infty$ with $m$ fixed to get $$\lim_{n \rightarrow \infty} a_n \geq \sum_{k=0}^{m}\frac1{k!}.$$
Taking the limit again as $m \rightarrow \infty$ we get $$\lim_{n\rightarrow \infty} a_n \geq e.$$
Therefore $$\lim_{n\rightarrow \infty} a_n = e.$$
