sequences and series - $limlimits_{ntoinfty} frac{n}{sqrt[n]{n!}} =e$

I don't know how to prove that
$$\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e.$$

Are there other different (nontrivial) nice limit that gives $e$ apart from this and the following
$$\sum_{k = 0}^\infty \frac{1}{k!} = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e\;?$$

Answer

In the series for $$e^n=\sum_{k=0}^\infty \frac{n^k}{k!},$$
the $n$th and biggest(!) of the (throughout positve) summands is $\frac{n^n}{n!}$.
On the other hand, all summands can be esimated as
$$\frac{n^k}{k!}\le \frac{n^n}{n!}$$
and especially those
with $k\ge 2n$ can be estimated
$$\frac{n^k}{k!}<\frac{n^{k}}{(2n)^{k-2n}\cdot n^{n}\cdot n!}=\frac{n^{n}}{n!}\cdot \frac1{2^{k-2n}}$$

and thus we find
$$\begin{align}\frac{n^n}{n!}Taking $n$th roots we find
$$\frac n{\sqrt[n]{n!}}\le e\le \sqrt[n]{2n+3}\cdot\frac n{\sqrt[n]{n!}}.$$
Because $\sqrt[n]{2n+3}\to 1$ as $n\to \infty$, we obtain $$\lim_{n\to\infty}\frac n{\sqrt[n]{n!}}=e$$
from squeezing.