Prove $3|n(n+1)(n+2)$ by induction

I tried proving inductively but I didn't really go anywhere. So I tried:

Let $3|n(n+1)(n+2)$.

Then $3|n^3 + 3n^2 + 2n \Longrightarrow 3|(n(n(n+3)) + 2)$

But then?

Here is a proof by induction.

The base case $n=0$ or $n=1$ is trivial.

Suppose that $3$ divides $n(n+1)(n+2)$ and consider $(n+1)(n+2)(n+3)$. Expand on $n+3$ and get $(n+1)(n+2)(n+3) = n(n+1)(n+2)+3(n+1)(n+2)$. Since by hypothesis $3$ divides the first term and it clearly divides the second one, it must divide the sum.

This proof can be generalized to $k \mid n(n+1)\cdots(n+k-1)$ using the same technique.

you can write $n=3\cdot k + r$ for some $k\in \mathbb{N}$ and $r\in \{0,1,2\}$. Then exactly one of $n,n+1,n+2$ is divisible by 3.

To do it inductively, we also need a base case: $0 \times 1 \times 2=0$ is divisible by $3$. Then, if $3$ divides $k(k+1)(k+2)=k^3+3k^2+2k$ then \begin{align*} (k+1)(k+2)(k+3) &= k^3+6k^2+11k+6 \\ &= (k^3+3k^2+2k)+3(k^2+3k+2). \end{align*} We know $k^3+3k^2+2k$ is divisible by $3$, by the inductive hypothesis, and $3(k^2+3k+2)$ is divisible by $3$ since $k^2+3k+2$ is an integer. Therefore, by induction, $k(k+1)(k+2)$ is divisible by $3$ for all $k \geq 0$.

Note, if we want to prove it for negative $k$ too, we'd need to perform "induction in reverse".