Is log n! = Θ(n log n)? Why is $\log(n!)=\Theta(n\log n)$?

I tried:
$\log(n!) = \log1 + \dots + \log n \leq n \log n \Rightarrow \log(n!) = O(n \log n)$.

But how can we prove $\log(n!) = \Omega(n \log n)$ without Sterling's approximation?
$$\begin{align*}
\log n! &amp;=\sum_{i=1}^n\log i \\
                   &amp;\geq \sum_{i=\lceil n/2\rceil}^n \log i \\
                   &amp;\geq \sum_{i=\lceil n/2\rceil}^n \log \tfrac{n}{2} \\
                   &amp;\geq \tfrac{n}{2}\log \tfrac{n}{2} \\
                   &amp;= \tfrac{n}{2}\big(\log n - \log 2)\\
                   &amp;= \Omega(n\log n)\,.
\end{align*}$$