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!}