TITLE: Proving that if $A^n = 0$, then $I - A$ is invertible and $(I - A)^{-1} = I + A + \cdots + A^{n-1}$
QUESTION [5 upvotes]: Let $A$ be a squared matrix, and suppose there exists an $n\in \Bbb N$ in a way that $A^n=0$. Show that $I-A$ is invertible and that $(I-A)^{-1}=I+A+\cdots+A^{n-1}$ 

I don't have a clue where to start from.

REPLY [6 votes]: How do you check a matrix $B$ is the inverse of matrix $A$? You multiply them together and see if you get the identity matrix. This is just the definition of inverse matrix.
You have a matrix $I-A$ and you want to check that its inverse is $1+A+A^2+...+A^{n-1}$ given that $A^n=0$.
So multiply together and after doing lots of cancellations etc you get $I - A^n = I$, which is what we were after!