Prove that $n! > \sqrt{n^n}, n \geq 3$ 
Problem Prove that $n! > \sqrt{n^n}, n \geq 3$. 

I'm currently have two ideas in mind, one is to use induction on $n$, two is to find $\displaystyle\lim_{n\to\infty}\dfrac{n!}{\sqrt{n^n}}$. However, both methods don't seem to get close to the answer. I wonder is there another method to prove this problem that I'm not aware of? Any suggestion would be greatly appreciated.

To show that $(n!)^2>n^n$ for all $n\geq 3$ by induction, you first check that $(3!)^2>3^3$.  Then to get the inductive step, it suffices to show that when $n\geq 3$, $(n+1)^2\geq\frac{(n+1)^{n+1}}{n^n}=(n+1)\left(1+\frac{1}{n}\right)^n$.  This is true, and in fact $\left(1+\frac{1}{n}\right)^n<3$ for all $n$.  It would be more than enough to use $\left(1+\frac{1}{n}\right)^n<n$, which is proved in another thread.
