The question is to prove that $$ (n!)^2 > n^n \text{ | } n\in\mathbb{N} $$

My attempt:

From the general AM-GM inequality, we have $$ \frac{a_1+a_2+a_3+\cdots+a_n}{n} \geq \sqrt[n]{a_1a_2a_3\cdots a_n} $$

Taking $$ a_r=\frac{1}{r^2} $$ we get

$$ \begin{aligned} &\left(\frac{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}{n}\right)^n > \left(\frac{1}{1.2^2.3^2\cdots n^2}\right) \\ \implies & \left(\frac{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}{n}\right)^n > \frac{1}{(n!)^2} \\ \implies & (n!)^2 > \left(\frac{n}{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}\right)^n \end{aligned} $$ If $$ \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)^n=k $$ Then, we have $$ (n!)^2 > \frac{n^n}{k} $$ and I'm stuck. Help please!

The question is to prove that $$ (n!)^2 \geq n^n \text{ | } n\in\mathbb{N} $$

My attempt:

From the general AM-GM inequality, we have $$ \frac{a_1+a_2+a_3+\cdots+a_n}{n} \geq \sqrt[n]{a_1a_2a_3\cdots a_n} $$

Taking $$ a_r=\frac{1}{r^2} $$ we get

$$ \begin{aligned} &\left(\frac{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}{n}\right)^n > \left(\frac{1}{1.2^2.3^2\cdots n^2}\right) \\ \implies & \left(\frac{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}{n}\right)^n > \frac{1}{(n!)^2} \\ \implies & (n!)^2 > \left(\frac{n}{1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}}\right)^n \end{aligned} $$ If $$ \left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)^n=k $$ Then, we have $$ (n!)^2 > \frac{n^n}{k} $$ and I'm stuck. Help please!