When looking through some old exercises, I have found one which stumped me a bit, and which has been bugging me for quite a few days. The question is how the polynomial $X^{p^n}-X-c$ factorizes over $\mathbb{F}_{p^n}$ (with $c \in \mathbb{F}_{p^n}$). The case when $n=1$ is easy, we find that the polynomial is irreducible. However in the more general case, it gets quite hard! I can't (I think) easily generalize the argument I used for when $n=1$, there I just used that if $r$ is a root of the polynomial in question we also find that $r \notin \mathbb{F}_{p}$ and that $r+i$ is a root for all $i \in \mathbb{F}_p$. The rest follows by looking at the explicit factorization over $\overline{\mathbb{F}_{p}}$
However when $n \neq 1$, I can't easily use the fact that $r \notin \mathbb{F}_{p^n}$. So my earlier arguments stops working, there's a hint that I could look at how the Frobenius automorphism acts on the roots of our polynomial, so i think I could use something about our polynomial being separable, however I didn't really see how. Any hints or answers would be greatly appreciated!
How about this approach? I hope that @JyrkiLahtonen will examine it closely, since I’m not completely convinced that all is well.
Call $p^n=q$, $k=\Bbb F_q$, $F(x)=x^q-x-c\in k[x]$, and $\varphi$ the $q$-Frobenius, sending $\xi\in\bar k$ to $\xi\,^q$. Let $\alpha$ be a root of $F$, i.e. of a $k$-irreducible factor $g$ of $F$. The $k$-conjugates of $\alpha$ are the finitely many $\varphi^m\alpha$, and the number of these is the degree of $g$. Now, $\varphi\alpha=\alpha^q=\alpha+c$, $\varphi^2\alpha=\varphi(\alpha+c)=\varphi\alpha+c=\alpha+2c$, and indeed $\varphi^n=\alpha+nc$ for any $n\in\Bbb Z$. Thus, if $c\ne0$, the number of $k$-conjugates of $\alpha$ is precisely $p$, and the irreducible factor $g$ of $F$ is of degree $p$.