Positive Integer Divison Proof Prove that $6|n(n + 1)(n + 2)$ for any integer $n ≥ 1$

I have attempted to do this but never seem to prove it.
As $3$ positive consecutive integers, either $n$, $n+1$, or $n+2$ must be divisible by $3$.

As $2$ positive consecutive integers, either $n$ or $n+1$ must be divisible by $2$.

Therefore, since $n(n + 1)(n + 2)$ is divisible by $2$ and $3$ ($2$ and $3$ are relatively prime numbers), it must be divisible by $6$.
If $n$ is even: $n = 2k \implies n(n+1)(n+2) = 2k(2k+1)(2k+2) = 2k(4k^2+6k+2)= 8k^3+12k^2+4k = 6k^3+12k^2 + 6k+2k^3-2k$. Observe that if $3 \nmid k\implies k^2-k = 0\pmod 3$ by Fermat Little's theorem. Thus $2k^3 - 2k = 0\pmod 6$. Thus $6 \mid n(n+1)(n+2)$. If $n$ is odd, then $n = 2k+1\implies n(n+1)(n+2) = (2k+1)(2k+2)(2k+3)= (m-1)m(m+1)$ with $2 \mid m = 2k+2$. So if $3 \nmid m\implies m^2-m = 0\pmod 3$ again by Fermat Little's theorem. Thus $6 \mid n(n+1)(n+2), \forall n \ge 1$.