I am not understanding a step in a proof from Rudin's Real and Complex Analysis. I am wondering why it is that $$\left[\frac{z^{n}-w^{n}}{z-w}-n w^{n-1}\right]=(z-w) \sum_{k=1}^{n-1} k w^{k-1} z^{n-k-1}$$ when $n\ge 2$. The details to obtain this equality are not apparent to me. Thank you!
10.6 Theorem If $f$ is representable by power series in $\Omega$, then $f \in H(\Omega)$ and $f^{\prime}$ is also representable by power series in $\Omega$. In fact, if $$ f(z)=\sum_{n=0}^{\infty} c_{n}(z-a)^{n} $$ for $z \in D(a ; r)$, then for these $z$ we also have $$ f^{\prime}(z)=\sum_{n=1}^{\infty} n c_{n}(z-a)^{n-1} $$ PROOF If the series $(1)$ converges in $D(a ; r)$, the root test shows that the series (2) also converges there. Take $a=0$, without loss of generality, denote the sum of the series $(2)$ by $g(z)$, fix $w \in D(a ; r)$, and choose $\rho$ so that $|w|<\rho<r .$ If $z \neq w$, we have $$ \frac{f(z)-f(w)}{z-w}-g(w)=\sum_{n=1}^{\infty} c_{n}\left[\frac{z^{n}-w^{n}}{z-w}-n w^{n-1}\right] $$ The expression in brackets is 0 if $n=1$, and is $$ (z-w) \sum_{k=1}^{n-1} k w^{k-1} z^{n-k-1} $$ if $n\ge 2$.
