Compound interest derivation of $e$ I'm reviewing stats and probability, including Poisson processes, and I came across:
$$e=\displaystyle \lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$
I'd like to understand this more fully, but so far I'm struggling. I guess what I'm trying to understand is how you prove that it converges. Can anyone point me toward (or provide) a good explanation of this?

It's not too hard to prove, but it does rely on a few things. (In particular the validity of the taylor expansion of $\ln$ around 1 and that $\exp$ is continuous.)
Consider in general the sequence $n\ln(1+x/n)$ which is defined for all $x$, positive or negative provided $n$ is large enough. (In fact the proof that follows can also be modified slightly to work for complex $x$). Using the power series representation $\ln(1+y) = \sum_{i=1}^\infty (-1)^{i+1}\frac{x^i}{i}$ we get
$$n\ln(1+\frac{x}{n}) = x + \sum_{i=2}^\infty(-1)^{i+1}\frac{x^i}{in^{i-1}}$$
The absolute values of the terms in the series on the right hand side are bounded above by a geometric series (for n large enough) whose summation tends to 0 as $n$ tends to infinity. Hence 
$$\lim_{n\rightarrow \infty} n\ln(1+\frac{x}{n}) = x$$
And so using the fact that $\exp(x)$ is a continuous function:
$$\lim_{n\rightarrow \infty}\bigg(1+\frac{x}{n}\bigg)^n = \exp\bigg(\lim_{n\rightarrow \infty}n\ln(1+\frac{x}{n})\bigg) = \exp(x)$$
In particular, set $x = 1$ to get the result you ask for.
