I cannot wrap my head around the $(n-1)$-th derivative of the polynomial $(z-2)^{n+1}$.
$$ \frac{d^{n-1}}{dz^{n-1}}(z-2)^{n+1} = \frac{(n+1)!}{2}(z-2)^2, \quad z \in \mathbb C.$$
I get why the term $(z-2)^2$ is there, the problem is with $(n+1)!/2$. Why divide by $2$?
The complex derivative works like the real one in this case: \begin{align*} \frac{d}{dz}(z-2)^{n+1} & =(n+1)(z-2)^n\\ (n+1)\frac{d}{dz}(z-2)^n & = (n+1)n(z-2)^{n-1} \\ & \vdots\\ (n+1)n\cdot\dots \cdot4\frac{d}{dz}(z-2)^3 & = \frac{(n+1)!}{2!}(z-2)^2 \\ \frac{(n+1)!}{2!}\frac{d}{dz}(z-2)^2 & = (n+1)!(z-2). \end{align*} Notice that the first derivative corresponds to having the exponent $n = (n+1)-1$ on $(z-2)$ on the right hand side of the first line. Thus at the $n$-th derivative you will have exponent $(n+1)-n=1$, meaning that the last line gives you the $n$-th derivative.
EDIT after the edit: for the $(n-1)$-th derivative just look at the line before the last.