Consider the field $\mathbb{F}_{q}$ for $q = p^k$. Let $f = X^p-X+a \in \mathbb{F}_q[X]$ with $a \in \mathbb{F}_q$, $a \neq 0$. I am trying to determine the degree of the irreducible factors of $f$ in terms of $a$ and $q$.
My strategy is to pick any root $\alpha$ of $f$, then this is a root of some irreducible factor $f_0 \ | \ f$, and this has degree $r$, where $r$ is the least natural number such that $\alpha^{q^r} = \alpha$. I found the following relation, inductively: $$ \alpha^{q^m} = \alpha + mka. $$ This equals $\alpha$ iff $mka \equiv 0 \mod q$, but how do I continu from here? I cannot determine what is the least $m$ for which this holds.
Let $\alpha$ be a root of $f$ in its splitting field, then $\alpha+\mathbb{F}_p$ is the set of roots of $f$, since $f$ is an affine linearized polynomial; i.e., $$f(\alpha+\beta)=f(\alpha)+\beta^p-\beta=f(\alpha)=0$$ for all $\beta\in\mathbb{F}_p$, so $$f(x)=\prod_{\beta\in\mathbb{F}_p}(x-\alpha-\beta).$$ Suppose that $f$ has a non-trivial factor $$g(x)=\prod_{i=1}^{m}(x-\alpha-\beta_i)$$ for some $\beta_i\in\mathbb{F}_p$. Then the coefficient of $x^{m-1}$ is $$-m\alpha-\sum_{i=1}^m\beta_i\in\mathbb{F}_q.$$ It follows that $\alpha\in\mathbb{F}_q$. That means if $f$ is reducible over $\mathbb{F}_q$, then it splits in $\mathbb{F}_q$.