I am given a square $n\times n$ matrix $A$ over an arbitrary field. For some natural number $m$ I have that $A^m = 0_V$. I want to prove that $I_n + A$ is invertible. I am not sure if my method covers all possibilities of what $A$ could be.
My method:
Letting $Q = I_n + A$. Then applying $A$, $m - 1$ times i.e. pre/post-multiplying by $A^{m-1}$ and re-arranging I get:
$$QA^{m-1} - A^m = A^{m-1}$$
Then simplifying
$$QA^{m-1} = A^{m-1}$$
Which then implies that $(Q-I_n)A^{m-1}=0_V$. I claim that this implies that $Q = I_n$ or $A^{m-1} = 0_V$ both of which imply $A = 0_V$ the latter by the above method. Which in turn show the invertibility of $Q$.
However it feels as if I am disregarding the possibility that $ZA^{m-1} = 0_V$ where $Z = Q-I_n$ and $Z$ and $A^{m-1}$ are both non-zero. Is this the case?
I also know I can prove invertibility by showing the matrix can be reduced to the identity matrix but using matrix algebra is far more approachable.