Given a nxn matrix $A$, show that if $A^3 = 0$ then $A-I_n$ is invertible. (Closed)

Question

So if we have $$A^3 = 0$$ we can say that $$A^3-I_n = -I_n$$ where $I_n$ is a n x n identity matrix. We know that to be invertible as all the columns of $I_n$ are linearly independent. I know we can factorize that to $$(A-I_n)(A^2+A+I_n)=-I_n$$ This is where I get stuck and am not sure on how to proceed.

Answer 1

Another proof: suppose that $A-I_n$ is not invertible. Then $0$ is an eigenvalue of $A-I_n$. Hence $1$ is an eigenvalue of $A$. It follows that $1=1^3$ is an eigenvalue of $A^3=0$, contradiction !