$\log (n!) $ Vs $(\log n)^{\log n}$

Question

$\log (n!)$ vs $(\log n)^{\log n}$

By Stirling approximation, $$\log n! = \Theta(n \log n)$$

Now comparing between $$ n \log n$$ vs $(\log n)^{\log n}$ {I am assuming $\log$ as base 2}

put $n=2^m$, we get $2^m \times m$ Vs $\log 2^{m \log 2^m} = m^2$ [Is it 1 or $m^2$?]

So, it is $m\times 2^m$ vs $m^2$

$\log (n!) < (\log n)^{\log n}$

Please tell me if this proof correct.

Answer 1

Since $(\log 2^m)^{\log 2^m}=m^m>m^2$, your conclusion is wrong.

Since $n!\le n^n$, $\log n!\le n\log n$ so$$\log\log n!\le\log n+\log\log n\le2\log n.$$But $\log[(\log n)^{\log n}]=\log n\times\log\log n$, which is greater for $n> b^{b^2}$ if the logarithms are in base $b$.