I have to show that $$\sum_{k=0}^\infty \frac{1}{k!}=e$$ using $$e=\lim_{n\to\infty }(1+\frac{1}{n})^n.$$
My attempt
$$(1+\frac{1}{n})^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{n!}{(n-k)!k!}\frac{1}{n^k}.$$ I have that $$\frac{n!}{(n-k)!}=\underbrace{(n-k+1)}_{\leq n}...\underbrace{(n-1)}_{\leq n}\underbrace{n}_{\leq n}\leq n^{k}\implies \frac{n!}{(n-k)!k!}\frac{1}{n^k}\leq \frac{1}{k!},$$ and thus $$(1+\frac{1}{n})^n\leq \sum_{k=0}^n\frac{1}{k!}\implies \lim_{n\to\infty }(1+\frac{1}{n})^n\leq \sum_{k=0}^\infty \frac{1}{k!}\implies e\leq \sum_{k=0}^\infty \frac{1}{k!}.$$ How can I do for the other inequality ?
Hint: It may be seen as a consequence of the dominated convergence theorem, since for every fixed $k$:
$$ \lim_{n\to +\infty}\frac{n!}{(n-k)!n^k} = \lim_{n\to +\infty}\left(1-\frac{1}{n}\right)\cdot\ldots\cdot\left(1-\frac{k-1}{n}\right) = 1.$$