Prove that $n(n^2 - 1) = \frac{(n+1)!}{(n-2)!}$

Question

Prove that for all $n \in \mathbb{N}$,

$$n(n^2 - 1) = \frac{(n+1)!}{(n-2)!}.$$

Thanks in advance.

Answer 1

$\require{cancel}$

If you really want to express $n(n^2 - 1)$ using factorials, note that $$n(n^2 - 1) = n(n+1)(n-1) = (n+1)(n)(n-1) = \frac{(n+1)n(n-1)\cancel{(n-2)}\cdots \cancel{2}\cdot 1}{\cancel{(n - 2)}\cancel{ (n-3)}\cdots \cancel{2}\cdot 1}$$

So we have that $$n(n^2 - 1) = \dfrac{(n+1)!}{(n-2)!}$$

So the given identity you are asked to prove is incorrect: the numerator needs to be $(n + 1)!$.

Answer 2

Here is how

$$n(n-1)=\frac{n(n-1)(n-2)!}{(n-2)!}=\frac{n!}{(n-2)!}$$

$$ n+1 = \frac{(n+1)!}{n!}.$$

Can you finish it?