Binomial series: why do we take $n \cdot (n-1)$ out?

Question

Given: $\sum_{k=1}^n \binom{n}{k} k(k-1)$. Find its closed form
We can multiply $(k^2-k)$ with $\binom{n}{k}$ and get:
$\frac{n!}{(k-2)!(n-k)!}$
Why do we take $n*(n-1)$ out: $n*(n-1)$*$\frac{(n-2)!}{(k-2)!(n-k)!}$

I dont understand this step: $\frac{n!}{(k-2)!(n-k)!}$ = $n*(n-1)$*$\frac{(n-2)!}{(k-2)!(n-k)!}$
Why do we do it?

Answer 1

The sum with $\dbinom nk$ is well-known and we try to reduce the sum to this form.

Using the factorial representation,

$$\binom nkk(k-1)=\frac{n!}{(n-k)!k!}k(k-1)=\frac{n!}{(n-k)!(k-2)!}=\frac{n(n-1)(n-2)!}{(n-k)!(k-2)!}=n(n-1)\binom{n-2}{k-2}.$$

Now $n(n-1)$ is constant and can be factored out of the sum and by shifting the indexes you can retrieve $\dbinom nk$.