Prove that $n^n \le (n!)^2$.

Question

Prove that $n^n \le (n!)^2$.

There is an elementary solution, which I haven't been able to find.

So far I tried manipulating and pairing terms but nothing worked.

I would appreciate any help!

Answer 1

Let $a_n=\frac{(n!)^2}{n^n}$, then $$r_n=\frac{a_{n+1}}{a_n}=(n+1)\big(\frac{n}{n+1}\big)^n\geq(n+1)e^{-1}$$

Since $n^n \le (n!)^2$ for $n=1,2$ and for $n\geq 2$, $r_n>1$ so we have the desired inequality for all $n$ by induction.

Answer 2

HINT:
Use the fact that $$ r (n+1-r) \geq n $$ (quadratic polynomial in $r$ with equality at $r=1$ and $r=n$) when $1\leq r\leq n$. Then, take the product over $r$: $$ \left(n! \right)^2=\prod_{r=1}^n r(n+1-r) \geq \prod_{r=1}^n n=n^n $$

Answer 3

Also you can use induction: $$ (n+1)^{n+1} = \frac{n^n}{n^n}{(n+1)}^{n+1}\le(n+1)(1+\frac{1}{n})^n(n!)^2\le{(n+1)^2(n!)^2}. $$