Why is $\sum_{k=0}^{n-1} (-1)^k a^{n-1-k}b^k = \left(\sum_{k=0}^{n-2} (-1)^k(k+1)a^{n-2-k}b^k\right)(a+b)+(-1)^{n-1}nb^{n-1}$?

Question

This transformation was used in answers to these questions:

$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$

Proving $\gcd( m,n)$=1 (the name is weirdly formulated, but those two are the same questions).

To someone who already asked the question I am asking, there was given a hint: $$a^p = ((a+b)-b)^p = \sum_{k=0}^p (-1)^k\binom{p}{k} (a+b)^{p-k}b^k.$$

Though, I wasn't able to use it either.