Are Z-transform time shifting and differentiation properties always compatible? I'm currently studying the Z-transform, and I'm having issues in understanding the time shift and differentiation properties, to be precise: calculating a Z-transform explicitly, and obtaining it by using those properties yields different results.

For example, a common Z-transform is the transform of the discrete unit step function:
$$
\mathcal{Z}\left\{ u\left[n\right]\right\} =\frac{1}{1-z^{-1}}=\frac{z}{z-1}
$$
The time shifting property of the Z-transform states (leaving aside its effects on the ROC for the moment) that $\mathcal{Z}\left\{ x\left[n-k\right]\right\} =z^{-k}\mathcal{Z}\left\{ x\left[n\right]\right\} \quad\forall k\in\mathbb{Z}
 $, thus it implies that:
$$
\mathcal{Z}\left\{ u\left[n-1\right]\right\} =z^{-1}\frac{z}{z-1}=\frac{1}{z-1}
 $$
Since $\frac{d}{dz}\frac{1}{z-1}=\frac{1}{\left(z-1\right)^{2}}$, the differentiation property, which states that $\mathcal{Z}\left\{ n\, x\left[n\right]\right\} =-z\frac{d\mathcal{Z}\left\{ x\left[n\right]\right\} }{dz}$ gives:
$$
\mathcal{Z}\left\{ n\, u\left[n-1\right]\right\} =-z\frac{1}{\left(z-1\right)^{2}}=-\frac{z^{-1}}{\left(1-z^{-1}\right)^{2}}
$$
which appears to be wrong, since the correct result should be:
$$
\mathcal{Z}\left\{ n\, u\left[n\right]\right\} =\frac{z^{-1}}{\left(1-z^{-1}\right)^{2}}
$$
What am I doing wrong? Is there perhaps any case in which those properties cannot be applied?
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.