TITLE: Prove that $n! > \sqrt{n^n}, n \geq 3$
QUESTION [17 upvotes]: 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.

REPLY [5 votes]: $\log$ is strictly concave $\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\log(x)=-\frac{1}{x^2}<0\right)$, so for $1\le k\le n$, we have
$$
\log(k)\ge\frac{(k-1)\log(n)+(n-k)\log(1)}{n-1}\tag{1}
$$
with equality only when $k=1$ or $k=n$. Summing $(1)$ yields
$$
\begin{align}
\log(n!)
&\ge\frac{n}{2}(\log(n)+\log(1))\\
&=\log(\sqrt{n^n})\tag{2}
\end{align}
$$
If $n\ge3$, then for at least one of the summands, equality fails; therefore,
$$
n!>\sqrt{n^n}\tag{3}
$$