Showing that $n\log n - n$ is $\Omega(n)$
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 $c$).