Using the binomial theorem:

\[ (x + h)^{n} = \sum_{i=0}^{n}\binom{n}{i}x^{n-i}h^{i} ,\]

we have:

\[ e^{x} = \lim_{n \to \infty} \sum_{k=0}^{\infty} \binom{n}{k} \frac{x}{n}^{k}. \]

In order to make sense of this, we have to find what

\[ \binom{n}{k} \left(\frac{1}{n}\right)^{k} = \frac{n!}{(n-k)!k!n^{k}} \]

becomes when \( n \) tends to infinity.

\[ % \lim_{n \to \infty} \frac{n!}{(n-k)!n^{k}k!} = \frac{1}{k!} \]

Note that:

\[ \frac{n!}{(n-k)!n^{k}} = \frac{n(n-1)...(n – k + 1)}{n(n)...(n)}. \]

This has k terms on top and k terms on the bottom.

Therefore:

\[ \lim_{n \to \infty} \frac{n!}{(n-k)!n^{k}} = \lim_{n \to \infty} \frac{n(n-1)...(n – k + 1)}{n(n)...(n)} = 1. \]

So that:

\[ \lim_{n \to \infty} \frac{n!}{(n-k)!n^{k}k!} = \frac{1}{k!}. \]

Putting this all together, we find:

\[ e^{x} = \lim_{n \to \infty} (1 + \frac{x}{n})^{n} \]

\[ = \sum_{k=0}^{\infty} \frac{x^{k}}{k!}. \]