I'm having a tough time deriving (4) from the bracketed expression in (3) shown in the photo. I've been futzing with partial sums of geometric series and binomial expansions for a while now with no luck. Anyone have an idea on this one? (This is from Rudin's Real and Complex Analysis btw...)
The expression on the right of (3) can be transformed into \begin{align} \frac{z^n-w^n}{z-w}-nw^{n-1}&=z^{n-1}\frac{1-\left(\frac{w}{z}\right)^{n}}{1-\frac{w}{z}}-nw^{n-1}=\\ &=z^{n-1}\left[1+\frac{w}{z}+\left(\frac{w}{z}\right)^{2}+\ldots+\left(\frac{w}{z}\right)^{n-1}\right]-nw^{n-1}=\\ &=\left(z^{n-1}+wz^{n-2}+w^2z^{n-3}+\ldots+w^{n-1}\right)-nw^{n-1}. \end{align} On the other hand, (4) may be rewritten as \begin{align} (z-w)\sum_{k=1}^{n-1}kw^{k-1}z^{n-k-1}&=\sum_{k=1}^{n-1}kw^{k-1}z^{n-k}-\sum_{k=1}^{n-1}kw^{k}z^{n-k-1}=\\ &=\sum_{k=1}^{n-1}kw^{k-1}z^{n-k}-\sum_{k=2}^{n}(k-1)w^{k-1}z^{n-k}=\\ &=\sum_{k=1}^{n-1}kw^{k-1}z^{n-k}-\sum_{k=1}^{n}(k-1)w^{k-1}z^{n-k}=\\ &=\sum_{k=1}^{n-1}w^{k-1}z^{n-k}-(n-1)\cdot w^{n-1}\cdot z^{n-n}=\\&=\sum_{k=1}^{n-1}w^{k-1}z^{n-k}-(n-1) w^{n-1}. \end{align} Can you see that both expressions coincide?