Showing that $n\log n - n$ is $\Omega(n)$ Prove that $n\log{n} − n$ is $\Omega(n)$.

I do know the answer:

$\log{n} ≥ 2, \forall n \ge 4$. 

Thus, $n\log{n}−n\ge n ,\forall n\ge 4 \implies n \log n − n ∈ Ω(n)$.

But can someone please explain how  its done. where did $2$, $\forall n$ and $4$ came from.
Then how was $n, \forall n \ge 4$ added.
According to the definition of big Omega, in order to show that $n\log n - n = \Omega(n)$, we need to come up with $n_0$ and $c$ such that all $n \geq n_0$ satisfy $n\log n - n \ge cn$.

Let us assume that the logarithm is to base 2. When $n \geq 4$, we have $\log n \geq \log 4 = 2$, and so $n\log n - n \geq 2n - n = n$. Therefore we can take $n_0 = 4$ and $c = 1$ and satisfy the requirements for $n\log n - n = \Omega(n)$.

The choice of $4$ is arbitrary &mdash; any $n_0 &gt; 2$ would work (for an appropriate value of $c$).