Compare the growth of the following 2 functions:
$$n^{\log \log \log n}$$ $$(\log n)!$$
My solution:
Let $n=2^m$
$$n^{\log \log \log n}=(2^m)^{\log \log \log_2 m}$$ $$=(2^m)^{\log \log m)}$$ $$=(\log m)^{\log 2^m}$$ $$=(\log m)^m$$
$$(log n)! = (log_2 2^m)!$$ $$= m!$$
Now I don't understand as to how to compare between
$$(\log m)^m$$ AND $$ m!$$
It would be a great if someone could help me with this. Thank you.