How does one get the following result $[z^n]\displaystyle\frac{z^k}{(1-z)^k} = {n-1 \choose k-1}?$
Where is the error in my computation?
$$\begin{align} [z^n]z^k\cdot (1-z)^{-k} &= [z^{n-k}](1-z)^{-k} \\ &=[z^{n-k}]\displaystyle\sum_{k \geq 0}{n + k - 1 \choose k }(-z)^k(1)^{-n-k}(-1)^k \\ &= [z^{n-k}]\sum_{k \geq 0}{n + k - 1 \choose k }z^k(-1)^{2k} \\ &= [z^{n-k}]\sum_{k \geq 0}{n + k - 1 \choose k }z^k\end{align}.$$
From here, I want the coefficient of $z^{n-k}$ in the sum, so I want that $k=n-k$ which happens when $n=2k$. But then something seems off here, i'm not even able to get the term with $z^{n-k}$ in the summation.
Change the index of summation to something other than $n$ or $k$. Then you get, replacing $k$ by $s$ as in the Wikipedia article,
$\begin{align} [z^n]z^s\cdot (1-z)^{-s} &= [z^{n-s}](1-z)^{-s} \\ &=[z^{n-s}]\displaystyle\sum_{k \geq 0}{s+k - 1 \choose k }(z)^s\\ &={s+n-s - 1 \choose n-s }\\ &={n - 1 \choose n-s }\\ &={n - 1 \choose (n-1)-(n-s) }\\ &={n - 1 \choose s-1 }\\ \end{align} $