Solving $\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$ I was solving a convergence of a series and this limit popped up:
$$\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$$
I needed this limit to be $0$ and it is in fact (according to WolframAlpha), but I just don't see how to get the result.

Consider that:
$$\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) = n, $$
hence:
$$ n! = \prod_{m=2}^{n} m = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}$$
and:
$$\frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{k}.\tag{1}$$
Since the sequence defined by:
$$ a_k = \left(1+\frac{1}{k}\right)^{k+\frac{1}{3}} $$
is increasing towards $e$, from $(1)$ it follows that:
$$ \frac{n^{n+\frac{1}{3}}}{n!}=\prod_{k=1}^{n-1} a_k \leq e^{n-1}\tag{2} $$
hence:
$$ \frac{n^n}{n!e^n}\leq\frac{1}{e\sqrt[3]{n}},\tag{3}$$
proving our claim.
