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 $e$?

Try bounding 
$$E_n = \left ( 1 + \frac{1}{n} \right )^n - \sum_{k=0}^n \frac{1}{k!}$$
using the binomial expansion you wrote, and pairing off terms. That is, write
$$E_n = (1-1) + (1-1) + (1/2 - 1/2n - 1/2) + \dots$$
Using the factorial definition of the binomial coefficient will prove useful.
