Confused about proof that $\log(n!) = \Theta(n \log n)$
Consider the sum $S = \log(1) + \dots + \log(n)$.  We're going to break it into two parts: $S=T+U$, where

\begin{align*}
T &amp;= \log(1) + \log(2) + \dots + \log(n/2)\\
U &amp;= \log(n/2+1) + \dots + \log(n-1) + \log(n).
\end{align*}

Basically, $T$ has the first $n/2$ terms of $S$, and $U$ has the remaining $n/2$ terms.

Now we'll lower-bound each of them.  Start with $T$.  Each term in $T$ is at least $\log(1)$, so we get

$$T \ge \log(1) + \log(1) + \dots + \log(1) = 0 + 0 + \dots 0,$$

so $T \ge 0$.  Next look at $U$.  Each term in $U$ is at least $\log(n/2)$, so we get

$$U \ge \log(n/2) + \log(n/2) + \dots + \log(n/2) = (n/2) \times \log(n/2).$$

Now $S = T+U$, so plugging in the lower bounds obtained above, it follows that

$$S =T+U\ge 0 + (n/2) \times \log(n/2).$$

This is exactly the result you wanted to prove.  Also, $(n/2) \times \log(n/2)$ is $\Omega(n \log n)$, so this proves that the sum $S$ is $\Omega(n \log n)$.