Consider the frobenius $\mathbb F_p^{alg}\longrightarrow \mathbb F_p^{alg}$ defined by $x\longmapsto x^p$.
1) Why is it surjective ? I recall that $\mathbb F_p^{alg}$ is an algebraic closure of $\mathbb F_p=\mathbb Z/p\mathbb Z$.
2) Can I then conclude that $x\longmapsto x^{p^n}$ is surjective ? I would say yes since if $\sigma :x\longmapsto x^p$, then $x\longmapsto x^{p^n}$ is in fact $\sigma ^n$ and thus it's the composition of $n$ surjection, and thus it's it self surjective.
In an algebraic closed field every non-constant polynomial has a root.
For a fixed $y$ a root $r$ of $X^p-y$ will satisfy $r^p=y$.