Complexity of $O(\log(n^n))$ vs $O(\log(n!))$
$O(\log n^n)$ and $O(\log n!)$ both describe the same set of functions.

Notice that $( \frac{n}{e} )^n \le n! \le n^n$ and hence:

$O(n \log n)  = O(n \log \frac{n}{e}) = O( \log ( \frac{n}{e} )^n) \subseteq O(\log n!) \subseteq O(\log n^n) = O(n \log n)$.

This shows that all the above sets coincide.