Prove the above by
by mathematical induction
By any other method.
I was just asked to prove this so I thought of using mathematical induction.
My effort :
I started first by verification and the inequality was true for n=3, 4...
Then the assumption step $(k!)^ 2 \gt k^k$
But after that while proving it for n=k+1 I was getting stucked inthe very next step.
Assuming that $(k!)^ 2 \gt k^k$ we write
$$((k+1)!)^ 2 = (k!)^2 \cdot (k+1)^2 \gt k^k \cdot (k+1)^2. \tag {*}$$
Now we compare the RHS (*) with $(k+1)^{k+1}\ $:
$$\dfrac{(k+1)^{k+1}}{ k^k \cdot (k+1)^2}=\dfrac{(k+1)^{k}}{ k^k \cdot (k+1) }=\dfrac{1}{k+1} \cdot {\left(1+\dfrac{1}{k}\right)^{k}}. $$
Now using the binomial formula, we have for $k \geqslant {3}$
$$
\left(1+\dfrac{1}{k}\right)^{k}=1+k\cdot \dfrac{1}{k}+\dfrac{k(k-1)}{2!\ k^2}+\ldots+ \dfrac{k(k-1)\cdots(k-(k-2)) }{(k-1)!\ k^{k-1}}+\dfrac{k(k-1)\cdots2\cdot{1}}{k!\ k^{k}}\\
< 1+1+\dfrac{1}{2!}+\ldots+\dfrac{1}{(k-1)!}+\dfrac{1}{k!} \\
< 1+1+\dfrac{1}{2}+\ldots+\dfrac{1}{2^{k-1}}+\dfrac{1}{2^k} < 3,
$$
thus for $k \geqslant {3}$
$$\dfrac{1}{k+1} \cdot {\left(1+\dfrac{1}{k}\right)^{k}} < 1.$$
Therefore,
$$ k^k \cdot (k+1)^2 > (k+1)^{k+1}.$$
Edit:
The multiplicative variant of Gauss's trick is very elegant. Writing squared factorial in the right and reverse order we have
$$(n!)^2 = [1\cdot 2 \cdots (n-1)\cdot n]\cdot[n\cdot (n-1) \cdots 2\cdot 1]= \prod_{k=1}^{n}{k(n+1-k)} . $$
In order to show that each term ${k(n+1-k)} \geqslant n$ we consider
$$k(n+1-k) - n = kn + k -k^2 - n = n(k-1) -k(k-1)=(k-1)(n-k) \geqslant 0$$
for $1 \leqslant k \leqslant n.$
Thus
$$\prod_{k=1}^{n}{k(n+1-k)} \geqslant \prod_{k=1}^{n}{n} = n^n.$$
