An element of $GL_n(\mathbb F_p)$ cannot have order $p^2$ if $n < p$

I'm preparing for my graduate program's entrance exams, and I came across this problem when studying. Our study group came up with a solution, but I wanted to ask if it was actually correct, since after reflection I think there is a big hole. I'm also curious if there is a solution that doesn't involve linear algebra.

The Problem

Let $G$ be the direct product of $n$ copies of $\mathbb Z_p$, where $n < p$. Show that no automorphism of $G$ has order $p^2$.

Solution

It is easy to see that the automorphism group of $G$ is isomorphic to $GL_n(\mathbb F_p)$, so it suffices to show that no matrix in $GL_n(\mathbb F_p)$ has order $p^2$. Let $M \in GL_n(\mathbb F_p)$ satisfy $M^{p^2} = I$. Then $M$ satisfies the polynomial $x^{p^2} - 1$, which over a field of characteristic $p$ is equal to $(x-1)^{p^2}$, and it follows that the eigenvalues of $M$ are all $1$. (This is the potentially problematic step: it might follow from the Cayley-Hamilton theorem and the fact that the characteristic polynomial of $M$ divides $x^{p^2} - 1$, but I'm not clear that the second thing is true...)

Now that we know that $M$ has only $1$ as an eigenvalue, consider the Jordan decomposition of $M$ over the algebraic closure $k$ of $\mathbb F_p$. It consists of Jordan blocks with (generalized) eigenvalue $1$ of size at most $n < p$, which means that the $p$th power of each block is the identity. Thus $M^p = I$ and $M$ cannot have order $p^2$.

Solution 1:

You can prove this without resorting to the Jordan form. Since $(X-1)^{p^2}$ kills $M$, you know that the minimal polynomial of $M$ is of the form $(X-1)^k$. Now the roots of $\chi_M$ are the same as those of the minimal polynomial, and $\deg \chi_M=n<p$, so $(X-1)^n$ kills $M$, and so does $(X-1)^p=X^p-1$.