Is $\log(n!)$ in $\Theta(n \log(n))$?
You mentioned
$$
\log(n!) = \log(n(n-1)\cdots1) = \log(n)+\log(n-1)+ \cdots +\log(1)
$$
From this, we can write (assuming $\log$ is base 2)
$$
\begin{array}{l}
\log(n!) \\
= \log(n)+\log(n-1)+....+\log(1) \\
\geq \log(n)+ \cdots + \log(n/2) \\
\geq \log(n/2)+ \cdots + \log(n/2) \\
= n/2 \cdot \log(n/2) \\
= n/2 \cdot (\log(n) - 1) \\
= n/2 \cdot \log(n) - n/2 \\
\end{array}
$$
since $\lim_{n\rightarrow +\infty} \dfrac{n/2}{n/2 \cdot \log(n)} = 0$, the last expression above is $\Theta(n \log(n))$, hence $\log(n!)$ is $\Omega(n \log(n))$.