Recurrence relations and the Master Theorem
We know that $n! \le n^n$, therefore $\log n! \le n \log n$. Then
$$
T(n) \le 99T( \frac{n}{100} ) + n \log n
$$
Let $c=\log_{100} 99 &lt; 1$ and notice that $n \log n$ is polynomially larger than $n^c$. Indeed:
$n \log n \in \Omega(n) \subset \Omega(n^c).$ By the master theorem we have $T(n) = O(n \log n)$.
This is tight because $\Omega(n \log n)$ is also a lower bound since:
$$
T(n) \ge \log n! \ge \log ( \lfloor n/2 \rfloor^{(n/2)} ) 
=  (n/2) \log ( \lfloor n/2 \rfloor)  = \Omega(n \log n).
$$
To summarize, you have $T(n) = \Theta(n \log n)$.