###### **Proof** Note that $x_{n}:= \left( 1+\frac{1}{n} \right)^{n} \leq e$: $\begin{align} x_{n} = \left( 1+\frac{1}{n} \right)^{n} &= \sum_{k=0}^{n} {n\choose k} \left( \frac{1}{n} \right)^{k} (1)^{n-k} = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} \frac{1}{n^{k}} \\ &= \sum_{k=0}^{n} \frac{1}{k!} \frac{(n)(n-1)\dots(n-k+1)}{n^{k}} \\ &= \sum_{k=0}^{n} \frac{1}{k!} \left( \frac{n}{n} \cdot \frac{n-1}{n}\dots \frac{n-k+1}{n} \right) \leq \sum_{k=0}^{n} \frac{1}{k!} \leq e \end{align}$ Similarly $\begin{align} \left( 1+\frac{1}{n} \right)^{m+n} &= \sum_{k=0}^{n+m} {n+m\choose k} \frac{1}{n^{k}} \\ &= \sum_{k=0}^{n+m} \frac{1}{k!} \frac{(n+m)!}{(n+m-k)!} \frac{1}{n^k} \\ &\geq \sum_{k=0}^{m} \frac{1}{k!} \frac{(n+m)!}{(n+m-k)!} \frac{1}{n^{k}} \\ &= \sum_{k=0}^{m} \frac{1}{k!} \frac{(n+m)(n+m-1)\dots(n+m-k+1)}{n^{k}} \\ & \geq \sum_{k=0}^{m} \frac{1}{k!} \end{align}$ Since $\frac{1}{\left( 1+\frac{1}{n} \right)}e\leq x_{n}\leq e$, by [[Sandwich Rule|sandwich rule]] $x_{n}\to e$ as $n\to \infty$.