$\lim\limits_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e$ 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
$$\sum_{k = 0}^\infty \frac{1}{k!} = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e\;?$$
 A: 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
$$\begin{align}\frac{n^n}{n!}<e^n&=\sum_{k=0}^{2n}\frac{n^k}{k!}+ \sum_{k=2n}^\infty \frac{n^k}{k!}\\&<(2n+1)\cdot\frac{n^n}{n!}+ \frac{n^n}{n!}\sum_{k=0}^\infty 2^{-k}\\&=(2n+3)\cdot\frac{n^n}{n!}.\end{align}$$
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.
A: I will use Cauchy-D'Alembert criterion
$$\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e$$
you can write the limit as $$\sqrt[n]{\frac{n^{n}}{n!}}.$$
Let be $\displaystyle x_{n}=\frac{n^{n}}{n!}.$
Now we can do $$\frac{x_{n+1}}{x_{n}}=\frac{\frac{(n+1)^{n}\cdot (n+1)}{n! \cdot (n+1)}}{\frac{n^n}{n!}}=\frac{(n+1)^{n}\cdot (n+1)}{n! \cdot (n+1)}\cdot \frac{n!}{n^{n}}=\frac{(n+1)^{n}}{n^n}=\left(\frac{n+1}{n}\right)^{n}=\left(1+\frac{1}{n}\right)^n \to e. $$
So $\displaystyle \lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e.$
A: Since 
$$
\ln \left( \frac{n}{\sqrt[n]{n!}}\right)=\ln n-\frac{\ln n!}{n}
$$
all you need is the weak Stirling formula:
$$
\ln n!=n\ln n -n +O(\ln n).
$$
By comparison with the integral of the nondecreasing function $\ln t$:
$$
\int_1^n\ln tdt\leq \ln n! =\sum_{k=2}^n \ln k\leq  \int_2^{n+1}\ln t dt.
$$
Recall that $\int \ln tdt =t\ln t -t +C$ and compute the lhs and the rhs.
The weak Stirling formula follows easily.
A: Use Stirling's approximation
$$\frac{n}{\sqrt[n]{n!}}\cong\frac{n}{\sqrt[n]{\sqrt{2\pi }\,n^{n+1/2}e^{-n}}}=\frac{1}{(2\pi)^{1/2n}}\frac{e}{n^{1/2n}}\xrightarrow[n\to\infty]{}1\cdot\frac{e}{1}=e$$
A: There are as many representations of $e$ as you could want at Wikipedia. $$e=\sum_1^{\infty}{k^7\over877k!}$$ $$e=\lim_{n\to\infty}n^{\pi(n)/n}$$ where $\pi(n)$ is the number of primes up to $n$. Any many, many more. 
A: I found this here:
$$e = \lim_{n \to \infty} \sqrt[n]{n\#}$$
Also see this related older post.
