 Solution

Interpret the displayed factorization as a proper factorization, with $1\le r\le p-1$; that restriction is necessary for the conclusion. In a splitting field, the monic degree-$r$ factor has $r$ of the distinct roots of the original polynomial. Comparing its coefficient of $X^{r-1}$ givesfor an $r$-element subset $J\subseteq {\mathbb{F}}_p$. Both $a_1$ and the final sum lie in $F$. Since $0<r<p$, the element $r$ is invertible in $F$, soPart (1) then gives complete splitting into distinct linear factors. Thus an Artin–Schreier polynomial of this form either is irreducible or splits completely; a trivial factorization with $r=0$ or $r=p$ would not imply a root in $F$.