Egorychev method how do you derive this geometric sum

Question

Example 2:

https://en.wikipedia.org/wiki/Egorychev_method#Example_II

How do you derive the closed-form result from the finite sum?

I tried it myself and the answer is way more complicated https://www.wolframalpha.com/input/?i=sum_%28k%3D1%29%5En++k*+%28z%2F%281-z%29%29%5Ek+

The leading 'k' value in the summation cannot just vanish like it does below? How does the k go away like that?

: \begin{align} & \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}} \sum_{k\ge 1} k \frac{z^k}{(1-z)^k} \; dz \\[6pt] = {} & \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}} \frac{z/(1-z)}{(1-z/(1-z))^2} \; dz \\[6pt] = {} & \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z^{n}} \frac{1}{(1-z)^n} \frac{1}{(1-2z)^2} \; dz. \end{align}

Answer 1

The $k$ goes away using the identity $$ \sum _{n=1}^\infty nx^n = \frac x {(1-x)^2}, $$ which is valid when $|x|<1$.