Prove the statement by the Principle of Mathematical Induction: $n^{3}-n$ is divisible by $6$ for each natural number $n \geq 2$.

(N/A) Let $P(n): n^{3}-n$ be divisible by $6$ for all $n \geq 2$.
Step $1$: For $n=2$,
$P(2): 2^{3}-2 = 8-2 = 6$,which is divisible by $6$.
Thus,$P(2)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \geq 2$,i.e.,$k^{3}-k = 6m$ for some integer $m$.
Step $3$: To prove $P(k+1)$ is true,we must show $(k+1)^{3}-(k+1)$ is divisible by $6$.
$(k+1)^{3}-(k+1) = (k^{3}+3k^{2}+3k+1) - k - 1$
$= (k^{3}-k) + 3k^{2}+3k$
$= (k^{3}-k) + 3k(k+1)$
Since $k^{3}-k = 6m$ and $k(k+1)$ is the product of two consecutive integers,it is divisible by $2$. Thus,$3k(k+1)$ is divisible by $3 \times 2 = 6$.
Therefore,$(k^{3}-k) + 3k(k+1) = 6m + 6n = 6(m+n)$,which is divisible by $6$.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \geq 2$.