Let $n = {1,2,3, ...}$, prove that $\left(1+\frac{1}{n}\right)^n$ is bounded above. I know that the sequence tends to $e \approx 2.718281828...$, but I am having a hard time proving it. Does anyone have any hints for me?

If $p=1,2,\cdots, n$, we have $(1-\frac{p}{n})<1$, and $2^{p-1}\le p!$, so $\frac{1}{p!}\le \frac{1}{2^{p-1}}$.
Therefore, if $n>1$, apply binomial theorem to $e_n$, which gives us
$$\begin{align}
e_n &= \left(1+\displaystyle\frac{1}{n}\right)^n\\
&= 1+1+\frac{1}{2!}(1-\frac{1}{n})+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+\cdots+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{n-1}{n})
\end{align}$$
 and use above two inequalities, we have $$2<e_n<1+1+\frac{1}{2}+\cdots+\frac{1}{2^{n-1}}=3-\frac{1}{2^{n-1}}<3$$
Hence $e_n$ is bounded above.
