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

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

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

Divide both sides by $n^{n-1}$ to arrive at $$ n>\left(1+\frac 1n\right)^{n-1}$$ to be shown. You may recognize that the right hand side converges to $e$, so we're in good shape. However, that is not explicit enough. So multiply with $(1-\frac1n)^{n-1}$ to get $$ n\cdot \left(1-\frac1n\right)^{n-1}>\left(1-\frac1{n^2}\right)^{n-1}$$ as goal. The right hand side is $<1$ for $n>1$. On the left hand side make use of Bernoulli's inequality $(1+x)^r\ge 1+rx$ if $x\ge-1$, $r\in\mathbb N_0$. So we have indeed $$ n\cdot \left(1-\frac1n\right)^{n-1}\ge n\cdot\left(1-\frac{n-1}n\right)=1>\left(1-\frac1{n^2}\right)^{n-1}.$$