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?