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.

$(n!)^2 = (n \times 1) \times ((n-1)\times 2) \times \cdots \times (1 \times n) \gt n^n$
since $(n-1)\times 2 = 2n-2 \gt n$ iff $n \gt 2$.  
Then take the square root. 
