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?

I have removed the more precise, but more confusing, justifications of $(1)$ and $(2)$. In this answer, I show that
$$
\left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n\tag{1}
$$
and
$$
\left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1}\tag{2}
$$
Therefore, $(1)$ says that $\left(1+\frac1n\right)^n$ is an increasing sequence and $(2)$ says that $\left(1+\frac1n\right)^{n+1}$ is a decreasing sequence. Since $\left(1+\frac1n\right)^n\le\left(1+\frac1n\right)^{n+1}$ the increasing sequence is bounded above and the decreasing sequence is bounded below; thus, they both converge.
Since $\lim\limits_{n\to\infty}\frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1n\right)^n}=\lim\limits_{n\to\infty}\left(1+\frac1n\right)=1$. They both converge to the same limit. This is called $e$.
$$
e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3}
$$
