TITLE: $x^p -x-c$ is irreducible over a field of characteristic $p$ if it has no root in the field
QUESTION [11 upvotes]: Let $c$ be an element of a field $F$ of characteristic $p$ (prime). Then how to show that $x^p -x-c$ is irreducible over $F$ if it has no root in $F$?

I was trying using contradiction and by considering the corresponding splitting field, but could not conclude.

REPLY [6 votes]: It's equivalent to show $x^p-x-c$ is reducible in $F[x]$ $\implies$ $x^p-x-c$ has a root in $F$.
Suppose first that $u$ is the root of $x^p-x-c$ in some extension field $K \supset F$, then $u, u+1, \cdots, u+p-1$ are all its roots.  
If $x^p-x-c$ is reducible in $F[x]$, then for $x-u, x-(u+1), \cdots, x-(u+p-1)$, there're $n \ge 1$ of them with their multiplication in $F[x]$.
The coefficient term of $x^{n-1}$ belongs to $F$, so $nu \in F$ and $u \in F$.