Does $A^{p^2}=I$ lead to $(A-I)^{p^2}=0$?

Question

Suppose that $A$ is a $n\times n$ matrix with entries in a finite field $F_p$ where $p$ is prime, $1\leq n \leq p$. An exercise answer told me that $$A^{p^2}=I\Rightarrow (A-I)^{p^2}=0.$$

I don't know why this holds. I tried to expand the equation as $$ 0=A^{p^2}-I^{p^2}=\left( A-I \right) \left( A^{p^2-1}+A^{p^2-2}+\cdots +A+I \right), $$ but still have no idea what info is given. Any help would be appreciated! Thank you in advance！

Answer 1

The binomial coefficients $\binom{p}{k}$ are all multiples of $p$ for $0 <k < p$. Therefore $$ (A-I)^{p^2}=((A-I)^p)^p=(A^p-I^p)^p=A^{p^2}-I=0 $$