How to show that $\log (n!) = \Omega(n\log n)$ using Stirling's approximation?
A function $f(n)$ is said be $\Omega(g(n))$ if there exist $c,n_0\gt 0$ such that $f(n)\ge cg(n)\ge 0$ for all $n\ge n_0$.

A function $f(n)$ is said be $\mathcal O(g(n))$ if there exist $c,n_0\gt 0$ such that $0\le f(n)\le cg(n)$ for all $n\ge n_0$.

Here,
$$\log_2(n!)=n\log_2 n-(\log_2 e)n+\mathcal O(\log_2 n)$$

Let $f(n)=\mathcal O(\log_2 n)$ such that
$$\log_2(n!)=n\log_2 n-(\log_2 e)n+f(n)$$

By definition, $f(n)\ge 0$.

Thus,
$$\begin{align}\log_2(n!)&amp;\ge n\log_2 n-(\log_2 e)n\end{align}$$

We want to find some constants $c,n_0\gt0$ such that
$$n\log_2n-(\log_2e)n\ge cn\log_2 n$$
for all $n\ge n_0$, that is, it must satisfy,
$$(1-c)n\log_2n\ge (\log_2e)n$$
As $n\gt 0$, it must satisfy
$$(1-c)\log_2n\ge (\log_2e)$$
Note that
$$(1-c)\log_2n\ge(1-c)\log_2n_0$$
for all $n\ge n_0$.
So, we need a positive $c$ and a $n_0$ to satisfy
$$(1-c)\log_2n_0\ge (\log_2e)$$
This is easily possible, as the RHS is a constant and the LHS can be made large by choosing $n_0$ and $c$ appropriately. For instance, choosing $c=0.1$ and $n_0=4$ works.

Thus, by definition,
$$\log(n!)=\Omega(n\log n)$$