As in the other solution, we consider only positive integers, as it is clear that $n^2 \ge -n$.
Our base case is $n = 1$: evidently, $1 \ge 1$, so this is true.
Now, our inductive step: we assume it is true for $n$ and try to prove it true for $n+1$. This means that we know that $n^2 \ge n$ and are trying to show that $(n+1)^2 \ge n+1$. Note that this is equivalent to showing $n^2 + 2n + 1 \ge n+1 \Rightarrow 2n \ge 0$, since we know $n^2 \ge n$. Now, it is evident that this is true for positive integers $n$, so we are done.
We could also just consider the following for positive integers $n$:
n^2 \ge n \Rightarrow n^2 - n \ge 0 \Rightarrow n(n-1) \ge 0
Noting that $n$ and $n-1$, if not $0$, are either both negative or positive, this expression is evidently true.