If $(A + I_n)^m = 0$, show that $A$ is invertible and find $\det(A)$

Question

Assume $(A + I_n)^m = 0$. Prove that $A$ is invertible and find $\det(A)$.

I started by binomial expansion, and set it equal zero. Is that correct? What would be the best approach?

Answer 1

The equation says that $A+I_n$ is nilpotent, so it has characteristic polynomial $(X+1)^n$. Setting $X=0$ in it gives $\det(-A)$, which apparently is $1$, so $\det(A)=(-1)^n$ (alternatively, $\det(A)$ is the product of all eigenvalues (counted with their algebraic multiplicity) and here they all are$~{-}1$).

Answer 2

Indeed, we can use the binomial expansion (since $A$ commutes with $I_n$) and we get $$ 0=\sum_{k=0}^m\binom mk A^kI_n^{m-k}=I_n+\sum_{k=1}^m\binom mk A^k =I_n+A\sum_{k=1}^m\binom mk A^{k-1} $$ hence we got the equality $$ \left(-\sum_{k=1}^m\binom mk A^{k-1}\right)\cdot A=I_n, $$ which shows that $A$ is invertible, and its inverse is $-\sum_{k=1}^m\binom mk A^{k-1}$.