What is Big-Omega notation of $\log(n!)$?

Question

I know that $\log(n!)$ is $O(n\log(n))$ since $n! < n^n$, but what is a lower bound for this function?

Answer 1

\begin{align} \log(n!) &= \log(n \cdot (n-1) \cdot (n-2) \cdot ... \cdot (n/2) \cdot (n/2 - 1) \cdot (n/2 - 2) \cdot ... \cdot 2 \cdot 1) \\\\ & \ge \log(n \cdot (n-1) \cdot (n-2) \cdot ... \cdot (n/2)) \\\\ & \ge \log(n/2 \cdot (n/2) \cdot (n/2) \cdot ... \cdot (n/2)) \\\\ & = \log( (n/2)^{n/2} ) \\\\ & = (n/2) \cdot \log (n/2) \\\\ & \ge cn\log(n) \text{ } \text{for some constant $c$} \end{align}