How can I prove the following Z-transform:
$$ Z\{n\} = \frac {z} {(z-1)^2} $$
As a tip, I was told to use the 'Multiplication in time'-property of the Z transform, which is the following:
$$ Z\{nx[n]\} = -z \frac{dX(z)}{dz} $$
But I don't see how I can use this.
Use the definition of $z$ transform
$$Z\left\{n\right\} = \sum_{n=0}nz^{-n}=\frac{z} {(z-1)^2}. $$
now you can use the geometric series (see my answer) to find the desired result.
Note: The $z$ transform of a sequence $a_n$ is given by
$$ \sum_{n=0}^{\infty} a_n z^{-n}.$$
I will show you, how you can prove it easily
$$u(n) \rightleftharpoons X(z) = \frac{z}{z-1}$$
According to the property which you mentioned
$$nu(n) \rightleftharpoons -z \frac{d(X(z)}{dz}$$
$$\frac{dX(z)}{dz} = - \frac{1}{(z-1)^2}$$
$$nu(n) \rightleftharpoons \frac{z}{(z-1)^2}$$
