Show that if $n>2$, then $(n!)^2>n^n$.

Show that if $n>2$, then $(n!)^2>n^n$.

My work:
I tried to apply induction.
So, at the induction step, I need to prove,
$n^n>(n+1)^{n-1}$
Here, I tried to use induction again without any luck. I also took log of both sides, but I did not get anything. Please help!

Solution 1:

Use a multiplicative variant of Gauss's trick: $$ (n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1) \ge n^n $$

Solution 2:

I think I saw a similar question here, but I can't remember where I saw it.

Here is the way:

$$(n!)^2=[1\times 2\times 3\times...\times n][1\times 2\times 3\times...\times n]$$ By grouping terms in pairs as in Gauss' trick, we write: $$(n!)^2=\prod_{i=1}^{n}i(n+1-i)$$ It's easy to see that $i(n+1-i)\geq n$ for every $i\in\{1,2,...,n\}$. Thus, we have: $$(n!)^2=\prod_{i=1}^{n}i(n+1-i)\geq n^n$$

I'll leave proving that we have a strict inequality for $n\geq 2$ to you