elementary number theory - Proof by induction that $n^3 - n$ is divisible by $6$

Show using induction that $n^3-n$ is divisible by 6 $\forall n\ge1, \quad n \in \mathbb{N}$

First off i show that the basis step: $1^3-1=0, \quad \frac{0}{6}=0$

Now I factorised it and set it equal to a multiple of 6: $\mathbf{n(n+1)}(n-1)=6A$

Assuming the result is true for k terms, and trying for $k+1$ terms:

$\mathbf{k(k+1)}(k+2)=6B$

I'm stuck here, I realise that the bold terms are the same, but $k+2$ and $n-1$ are not. Could someone show me what do to next to solve this.

Also is it possible to prove this using modular arithmetic?

Thanks,