Are Z-transform time shifting and differentiation properties always compatible?
First

$$\frac{d}{dz}\frac{1}{z-1}=-\frac{1}{(z-1)^2}$$

(with a minus sign in front). So you have

$$\mathcal{Z}\{nu[n-1]\}=z\frac{1}{(z-1)^2}=\frac{z^{-1}}{(1-z^{-1})^2}$$

Now everything is OK because

$$nu[n-1]=nu[n]$$

because for $n=0$ also the right-hand side vanishes, and for $n&gt;0$ both are equal anyway.