Here is the problem:
Suppose $f(x) \in k[x]$ is monic and has distinct roots in some splitting field. If these roots form a field then show that the characteristic of $k$ is $p$ and that $f(x) = x^{p^n} - x$.
What I basically did I said let $F = \{r : f(r) = 0\}$, where $r$ is a root in this splitting field. Then $F$ is a finite field, by hypothesis so the characteristic of $F$ is $p$ some prime, and $F^\times$ is cyclic. In particular, $|F^\times| = p^n - 1$, for some integer $n \geq 1$. By Lagrange theorem, $x \in F$ satisfies $|x| \mid |F^\times|$which after a bit of manipulation shows you that $x^{p^n} - x = 0$. Since every root of $f$ is in $F$, this means $f(x) = x^{p^n} - x$, as desired.
Questions What I'm strugling to understand is (1) where we used that the roots were attained in the splitting field. It seems like I could have just replaced this with algebraic closure and been fine, but then the result seems absurd. So it is possible that my proof is wrong or I'm misunderstanding something. (2) I also don't know how to show the second part that the characteristic of $k$ is $p$? Hint here would be great.
The splitting field is not important here at all; it's just a place for the roots to live so we can talk about them. The argument works perfectly well if you instead take the roots of $f$ in any field over which $f$ splits, including an algebraic closure of $k$. I don't know what you think is "absurd" about this.
Well, you know the characteristic of $F$ is $p$, and $k$ and $F$ are both subfields of a common field (the splitting field of $f$ over $k$).