$$\frac{n^{\ln n}}{(\ln n)^n} < \frac{1}{n^2} $$ for $n \ge 13, n \in N$
I did some transformations:
$n^2 \times n^{\ln n} < (\ln n)^n $
$n^{2 + \ln n} < (\ln n)^n$
$\log n^{2 + \ln n} < \log (\ln n)^n$
$(2 + \ln n)\log n < n\log (\ln n)$
But I don't know what I should do next.
We have that
$$\frac{n^{\ln n}}{(\ln n)^n} < \frac{1}{n^2} \iff {\ln^2 n+ 2\ln n-n \ln(\ln n)} <0$$
and
$$f(x)=\ln^2 x+ 2\ln x-x \ln(\ln x) $$
$$\implies f'(x)=\frac{2\ln x}{x}+\frac 2 x-\ln(\ln x)-\frac1 {\ln x}$$ $$\implies f''(x)=\frac{x-2\log^3 x-x\log x}{x^2\ln^2 x}$$
and since $f''(x)<0$ as $x>e$ we have that $f'(x)$ is strictly decreasing as $x>e$ and since $f'(5)<0$ $f(x)$ is strictly decreasing as $x>5$ and since $f(13)<0$ we can conlude that the inequality holds for any $n\ge 13$.