Complexity of $O(\log(n^n))$ vs $O(\log(n!))$ Is $O(\log(n^n)) &lt; O(\log(n!))$? Is there any good/practical algorithm with this kind of complexity? 

And also, to check my understanding of algorithmic complexity, are these two $&gt; O(n\log(n))$?

I'm just starting with this topic and not really sure if I'm getting it right.

Thanks and stay safe.
$O(\log n^n)$ 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.