Prove that $6$ divides $n(n + 1)(n + 2)$

$n$ is either even or odd. Also, it must be congruent, modulo $3$, to either $0$, $1$, or $2$.

Another possibility:

$$\frac{n(n+1)(n+2)}{6}$$ equals the binomial coefficient $$\binom{n+2}{3}$$ which is an integer (since it is the number of 3-element subsets of an ($n+2$)-element set).

The links provided by lab bhattacharjee solve the problem in more generality, but for the sake of completeness...

Let us do an inductive proof. For $n = 0$ things are trivial. Suppose we know the claim for $n$, and we want to prove it for $n+1$. By inductive assumption, $$n(n+1)(n+2) = n^3+2n^2 + 2n = 6 q$$ for some $q$. Now, $$(n+1)(n+2)(n+3) = n(n+1)(n+2) + 3(n+1)(n+2) = 6q + 3(n+1)(n+2)$$ A simple argument (use induction again, if you like) shows that $(n+1)(n+2)$ is divisible by $2$, so you can write $(n+1)(n+2) = 2p$. Consequently, $$(n+1)(n+2)(n+3) = 6(q+p)$$ which finishes the problem.

The product $n(n+1)(n+2) \equiv 0\ (\mathrm{mod}\ 2)$, since if $n$ is even the product will be even, and if $n$ is odd then $n+1$ will be even and again the product is even.

We also know that $n(n+1)(n+2) \equiv 0\ (\mathrm{mod}\ 3)$. You can prove this by notating that $n\ \mathrm{mod}\ 3$ must be $0, 1,$ or $2$, and all of these cases imply the product is divisible by $3$. $$n\equiv0\ (\mathrm{mod}\ 3)\implies n(n+1)(n+2)\equiv0\ (\mathrm{mod}\ 3),$$ $$n \equiv1\ (\mathrm{mod}\ 3)\implies n+2\equiv 0\ (\mathrm{mod}\ 3)\implies n(n+1)(n+2)\equiv0\ (\mathrm{mod}\ 3),$$ $$n\equiv2\ (\mathrm{mod}\ 3)\implies n+1 \equiv 0\ (\mathrm{mod}\ 3) \implies n(n+1)(n+2)\equiv0\ (\mathrm{mod}\ 3).$$

Therefore $n(n+1)(n+2)$ is divisible by $2$ and by $3$, which implies it is divisible by $6$.