How to show that $n^2<(\ln n)^n$ holds when $n$ is sufficiently large?

Question

I am struggling to prove that there exists $N\in\mathbb N$ so that for all $n>N$, $n^2<(\ln n)^n$ holds.

By inspection, I found out that for all $n>6$, $n^2<(\ln n)^n$. I don't know how to prove it formally. What are useful identities I can use?

Answer 1

Let $f(x)=2\ln{x}-x\ln\ln{x}.$

Thus, for $x>6$ $$f'(x)=\frac{2}{x}-\ln\ln{x}-\frac{1}{\ln{x}}<\frac{1}{3}-\ln\ln6<0,$$ $f'$ decreases and $$\lim_{x\rightarrow+\infty}f'(x)=-\infty,$$ which says $\lim\limits_{x\rightarrow+\infty}f(x)=-\infty.$

Answer 2

Apply $\ln$ (strictly increasing function) to get $\ln n^{2}<\ln\left(\ln n\right)^{n}$ or $2\ln n<n\ln\left(\ln n\right)$.

For all $n$, $n>\ln n$.

For all $n\gtrsim1619$, $\ln\left(\ln n\right)>2$ and so $2\ln n<n\ln\left(\ln n\right)$.