How to prove that $n^n=O(\log_2(2^{2^{2n}}))$?

Question

How to prove that $n^n=O(\log_2(2^{2^{2n}}))$?

I cant find the the constant that will make it true.

Thanks for the help

Answer 1

@NN2 is right. Note that $\log_2\left(2^{2^{2n}}\right)=2^{2n}=4^n\in o(n^n)$. On the other hand,$$n^n=2^{n\log_2n}\ll2^{n^2}\ll2^{2^{kn}}=\log_2\left(2^{2^{2^{kn}}}\right),\,k>0,$$so we can fix the problem statement by replacing the innermost $2n$ with $2^n$ or $2^{2n}$. The latter effect is also achieved with an extra $2^{\cdot}$ operation around the argument of $\log_2$, or equivalently deleting the $\log_2$.