Finding $\lim_{n\to\infty}\frac{2^n\cdot n!}{n^n}$ 
$\lim_{n\to\infty}\frac{2^n\cdot n!}{n^n}$

I think we can do it by squeeze rule but I could not find a sequence bigger than this one converging to 0 without using Stirling approximation of n!

Being a sequence of positive terms you can use D'Alambert's criterion.
$$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}\frac{n^n}{2^n n!}=2\left(\frac{n}{n+1}\right)^n\longrightarrow 2e^{-1}<1$$
and so $a_n=\frac{2^n n!}{n^n}\longrightarrow0$.
