## How to show that $n^{\log_2(\log_2(n))} \in O(2^n)$?

In order to show that $$f(n) = n^{\log_2(\log_2(n))} \in O(2^n)$$ we need to find constants $$n_0 > 0$$ and $$c > 0$$ such that for all $$n \ge n_0$$:

\begin{align} 0 \le n^{\log_2(\log_2(n))} \le c2^n. \end{align}

If we choose $$n_0 = 4$$, then $$\log_2(\log_2(4)) = 1$$ and so for all $$n \ge n_0: f(n) \ge n > 0$$.

My question is which inequalities can I use to make it obvious that $$n^{\text{log}_2(\text{log}_2(n))} \le 1*2^n$$, for $$c = 1$$ and all $$n \ge 4$$?

How do I have to reason in general to give a good upper bound for a composition of multiple $$\log_b$$ functions?