Showing that the following function is O of the other

Question

I have the following functions: $f_1= $ $n\choose{0.5n}$ and $f_2 = (\log\log n)^{\log n}$. Im trying to prove that $f_2=O(f_1)$. I simplified $f_2$ to the following: $f_2 = (\log\log n)^{\log n}=2^{\log((\log(\log(n)){}^{\log n}))}=2^{\log(n)*\log((\log(\log(n))))}$ and approximated $f_1$ using Sterlings approximation to the following:$\frac{2^{n}}{\sqrt{n}}\leq f_{1}$. so, I'm left with showing that: $2^{\log(n)*\log((\log(\log(n))))}\leq \frac{2^{n}}{\sqrt{n}}$. I tried to prove that $2^{\log(n)*\log((\log(\log(n))))}\leq 2^{0.5n}$ but cant seem to make progress. Help is much appreciated. thanks in advance!

Answer 1

The trick was using the fact that $log(n)=O(n^\epsilon)$.