logarithmic big O proof

Question

I am asked to prove that $log^{log n}log n$ = O($2^{\sqrt n})$ I've tried so much and don't have any idea to solve it please help thank you.

Answer 1

$$f(n)=\log^{\log(n)}\log(n)$$

$$\log[f(n)]=\log(n)\log[\log(\log(n))]$$

and for $n>100$.

$$<\sqrt n$$

Thus, $\log[f(n)]=\mathcal O(\sqrt n)\implies f(n)=\mathcal O(a^{\sqrt n})$