Show that $ (\mathbf{1}_n-A)^{-1} = \displaystyle \sum_{l=0}^m A^l $ , how to approach that problem?

Question

Let $ A \in \mathbb{C}^{n \times n} $ be nilpotent, i.e. there is a $ m \in \mathbb{N} $, such that $ A^{m} $= 0

Show that $$ (\mathbf{1}_n-A)^{-1} = \displaystyle \sum_{l=0}^m A^l $$

I am fully missing the propositions here to transform the LHS such that the RHS occurs. Do anyone have an approach or a tip or the propositions themselves?

Answer 1

Hint: Two matrices are inverses if and only if they multiply to the identity. Can you find a way to simplify the expression $$(I-A)\left(\sum_{\ell=0}^m A^\ell\right)$$ to show that it is the identity?