$$ \frac{1}{(\omega - z)^n} - \frac{1}{\omega^n} = \frac{z}{(\omega - z)^n\omega^n} \sum\limits_{k=0}^{n-1}\omega^{n-k-1}(\omega-z)^k $$
I multiplied by $(\omega - z)^n\omega^n$ and arrived at the following equality I am struggling to prove
$$\sum\limits_{k=0}^{n-1}(-1)^k {n\choose k}\omega^{n-k-1}z^k = \sum\limits_{k=0}^{n-1}\omega^{n-k-1}(\omega-z)^k.$$
Since the first equation is an (apparently) easy step in a prove it shouldn't be too difficult.
Use the identity: $\displaystyle a^n-b^n = (a-b)\left(\sum\limits_{k=0}^{n-1}a^kb^{n-k-1}\right)$ with $a = w$ and $b = w -z$
You can divide both sides by $b^n$ and identify it as a GP-sum in $a/b$.