discrete mathematics - Induction Proof: $2$ divides $n^2 + n$ for each $n in mathbb{N}$

So I am looking at some induction questions and I am trying to solve them on my own but I am getting stumped and frustrated. There was a previous question question that was answered, but I changed it to see if I could solve it. I am not getting that far.

How do I show by mathematical induction that $2$ divides $n^2+n$ for all $n$ belonging to the set of Natural Numbers. Here is what I have so far. Could I be pointed in the right direction? You can see below where I am stumped. This is where I am having the issue.

Basis: $n=1, \qquad P(1)$ is true, 2 divides $1^2+1 = 2$

Induction Hypothesis: 2 divides $(k+1)^2+(k+1)$ for some $k \in \mathbb{Z} \geq 1$

Induction Step: $(k+1)^2+(k+1)=k^3+3k^2+3k+1=$

Answer

Alright, so our inductive hypothesis is that $k^2 + k$ is a multiple of $2$ for some $k$. Then consider $(k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = (k^2 + k) + 2k + 2 = (k^2 + k) + 2(k + 1)$.

By our inductive hypothesis, $k^2 + k$ was an even number: because we are adding an even number, $2(k+1)$, to it, we still have an even number.

Therefore, $(k+1)^2 + (k+1)$ must be even as well. This completes our proof.