Using the Well Ordering Principle to Show that n^3 - n is divisible by 6. A Guide On Well-Ordering Principle

MIT OCW Guide on WOP

I read the following two sources and I think I understand the structure of proofs involving WOP, but I am stuck on the open-ended part where you prove P(n-1) for the  following problem: 


Use the well-ordering principle to show that n^3 - n is divisible by 6. 


Any help would be greatly appreciated! 

Note: After factoring out (n-1)^3 - (n-1) it took the form of n^3 - 3n^2 + 2n and the only difference between that and n^3 - n is "3n^2 - 3n" which I proved was divisible by 6 via induction. Is there a better way to solve this problem?

Edit: Show that $n^3-n$ is divisible by $6$ using induction, post is useful however I not supposed to use induction for the main part of the proof.
The basic proof strategy using the WOP is that you need to show that there are no exceptions to the claim that:

$P(n): n^3-n$ is divisible by $6$

and you do this by a proof by contradiction:

Assume there are exceptions. Then let $k$ be the smallest number for which the claim does not hold, i.e. assume that $\neg P(k)$.  Now, if we can now show that $\neg P(k-1)$, then we have our contradiction, because $k$ was supposed to be the smallest number for which the claim did not hold.  That is, you need to show that:

$\neg P(k) \Rightarrow \neg P(k-1)$

(Note: this assumes $k&gt;0$, or, if you start with $1$, that $k&gt;1$. ... if $k=0$ (or $k=1$), you show there is a contradiction by showing that in fact we do have $P(0)$.  So note how WOP is really equivalent to induction, where you show $P(0)$, as well as $P(k-1)\Rightarrow P(k)$, which of course is just the contrapositive of $\neg P(k) \Rightarrow \neg P(k-1)$)

Well, you can show that $(k-1)^3-(k-1)=(k^3-k)-(3k^2-3k)$

So, given the assumption that $\neg P(k)$, i.e. that $(k^3-k)$ is not divisible by $6$, then (start of inside proof by contraposition) if $(k-1)^3-(k-1)$ were to be divisible by $6$, then $(3k^2-3k)$ cannot be divisible by $6$. OK, but obviously $(3k^2-3k)=3(k^2-k)$ is divisible by $3$. And for the remaining term $(k^2-k)$ we have that $(k^2-k)=k(k-1)$, and obviously either $k$ or $k-1$ is even, and so that term is divisible by $2$, and so $(3k^2-3k)$ is divisible by $6$ after all: Contradiction! So, $(k-1)^3-(k-1)$ is not divisible by $6$, under the assumption that $(k^3-k)$ is not divisible by $6$
$n^3-n=n(n-1)(n+1)$ which is product of three consecutive numbers. 

Well if you have three consecutive numbers one of them is a multiple of two and one is a multiple of three so the product is a multiple of six.