I am working on the following problem:
Let $k$ be a field of characteristic $p > 0$. Let $f(x) = x^p - x - a$ with any element $a \in k$. Suppose that $f(x)$ has a root in $k$. Show that $f(x)$ has exactly $p$ roots in $k$.
First, I considered the field $\mathbb{F}_p$ , which is a field with characteristic $p$. Let $\alpha$ be a root of $f(x) = x^p - x - a$ in $\mathbb{F}_p$. Then $\alpha + 1$ is a root of $f(x)$ in $\mathbb{F}_p$, since $f(\alpha + 1) = (\alpha + 1)^p - (\alpha + 1) - a = \alpha^p + 1^p - \alpha - 1 - a = \alpha^p - \alpha - a = 0$. If we continue this inductively $p$ times, we find that $\alpha + \beta$ ($1 \leq \beta \leq p$) are all roots of $f(x)$ in $\mathbb{F}_p$. Thus, every element of $\mathbb{F}_p$ is a root of $f(x)$, giving exactly $p$ roots in $\mathbb{F}_p$ as desired.
Now, I'm struggling with generalizing this to the field $GF(p^n)$, $n > 1$, since this is also a field of characteristic $p$. The same strategy above will show that one can obtain $p$ roots of $f(x)$ in this field -- but how can I show that there are no more roots beyond those $p$ ?
Thanks!