How to show that $ ({\bf I_n}-A)^{-1} = \Sigma_{l=0}^m A^l $ I struggle with the following task:

Let $n\in\mathbb{N}$ and $A\in\mathbb{C}^{nxn}$ be nilpotent, i.e. there is $m\in\mathbb{N}$ such that  $A^m$ = 0
Show that $ ({\bf I_n}-A)^{-1} = \Sigma_{l=0}^m A^l $

Can anyone help me out?

A nilpotent matrix $A$ only has eigenvalues which are $0$.
The matrix $I-A$ then only has eigenvalues $1$. $(I-A)^{-1}$ will have inverted eigenvalues and $1^{-1}=1$
What does Caley-Hamilton theorem tell us? $(I+A-I)^k =0 \Leftrightarrow A^k=0$.
But we may already know this from nilpotency.
Another fruitful aspect of this is the geometric series. Maybe this question helps you there.
