How do I prove that $X^{p^n}-X$ is the product of all monic irreducible polynomials in $\mathbb Z_p[X]$ of degree dividing $n$?
Let $\bar Z_p$ be an algebraic closure of $Z_p$.
Define $F=\{x\in \bar Z_p|x^{p^n}-x=0\}$.
Then $X^{p^n}-X=\prod_{\alpha\in F} X-\alpha$.
Now $S$ be the set of monic prime polynomials of $Z_p[X]$ of which degree divides $n$.
Then, I know that $F=\bigcup_{f\in S} \{x\in \bar Z_p| f(x)=0\}$.
With this information, how do I prove that $X^{p^n}-X$ is actually the product of all elements of $S$?
To assure the equality, I think it must be shown that $\{x\in \bar Z_p|f(x)=0\}$ are mutually disjoint, but I don't know how to show this
Some ideas:
Prove that $\;w\;$ is a multiple root of a non-zero polynomial $\;f\;$ iff $\;w\;$ is also a root of its derivative $\;f'\;$. Deduce that if $\;f\;$ is irreducible (over some field), then this happens iff $\;f'=0\;$, and thus over a field of characteristic $\;p>0\;$ this can happen iff all the non-zero coefficients of $\;f\;$ correspond to powers of $\;x\;$ which are multiples of $\;p\;$.
Thus, $\;T(x):=x^{p^n}-x\in\Bbb F_p[x]\;$ is separable, meaning: all its roots are simple.
Now, using your notation, show that $\;\Bbb F_{p^n}=\{w\in\overline{\Bbb F_p}\;:\;\;T(w)=0\}$ , and finally: show that $\;\Bbb F_{p^m}\le\Bbb F_{p^n}\iff m\mid n\;$ .