$M^n-I$ invertible implies $M$ nilpotent over finite field Let $M$ be a matrix over the finite field $\mathbb F_p$ ($p$ prime) such that $M^n-I$ is invertible for all positive integers $n$. Prove that there exists $k$ such that $M^k=0$ (i.e. $M$ is nilpotent).
The condition is equivalent to $\det(M^n-I)\neq0$ for all $n$. So using $$M^{2n}-I=(M^n-I)(M^n+I),$$ I have shown that $M^n+I$ must always be invertible too.
It would be enough to show that the only eigenvalue of $M$ is $0$, but I'm not sure how to do that.

There are only finitely many matrices in $\mathbb F_p$ with the same size as $M$, so the list $M, M^2, M^3,\dots$ repeats eventually, say $M^k=M^\ell$ for $k<\ell$. Then $M^k\left(M^{\ell-k}-I\right)=0\implies M^k=0$, as $M^{\ell-k}-I$ is invertible.
