Prove that $6$ divides $n(n + 1)(n + 2)$ I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows:

Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$.
Note: An integer $a$ divides an integer $b$, written $a|b$, if there exists $q ∈ Z$ such that $b = qa$. Alternatively, $a|b$ if dividing $b$ by $a$, $b ÷ a$, results in an integer.

Should I do a proof by induction?
All help/input is appreciated!

If we let $m=n+1$, we need to prove that $6 \vert m(m^2-1)$ for all $m \geq 2$. Equivalently, we want to show that
$$m^3 \equiv m \bmod6$$
First note that $m \equiv 0,1,2,3,4,5 \pmod 6$. Now observe that following:
$$m \equiv 0 \bmod 6 \implies m^3 \equiv 0^3 \bmod 6 \implies m^3 \equiv 0 \bmod6 $$
$$m \equiv 1 \bmod 6 \implies m^3 \equiv 1^3 \bmod 6 \implies m^3 \equiv 1 \bmod6$$
$$m \equiv 2 \bmod 6 \implies m^3 \equiv 2^3 \bmod 6 \implies m^3 \equiv 8 \bmod6 \implies m^3 \equiv 2 \bmod6$$
$$m \equiv 3 \bmod 6 \implies m^3 \equiv 3^3 \bmod 6 \implies m^3 \equiv 27 \bmod6 \implies m^3 \equiv 3 \bmod6$$
$$m \equiv 4 \bmod 6 \implies m^3 \equiv 4^3 \bmod 6 \implies m^3 \equiv 64 \bmod6 \implies m^3 \equiv 4 \bmod6$$
$$m \equiv 5 \bmod 6 \implies m^3 \equiv 5^3 \bmod 6 \implies m^3 \equiv 125 \bmod6 \implies m^3 \equiv 5 \bmod6$$
Hence, in all cases, we have $m^3 \equiv m \bmod 6$, which means that $$6 \vert (m^3-m) \implies 6 \vert n(n+1)(n+2)$$
