A polynomial's irreducibility in $\Bbb{Z}_p$ 
Show that if $f$ is irreducible in $\Bbb{Z}_p[x]$ then $f$ divides $x^{p^n} - x$ for some $n \in N$.

I know that: 


*

*$f$ is irreducible, so $F = \Bbb{Z}_p / {\left\langle f\right\rangle}$ is a finite field of size $p^m$ for some $m$.

*Every element of of $F$ satisfies $x^{p^m} = x$.

*$f$ is solved in $F$ by some element $x / \left \langle f \right \rangle$.


Just a hint on how to tie it together would be helpful. I feel like I am missing something obvious.
 A: Let $u = x + \langle f\rangle \in \Bbb Z_p[x]/\langle f\rangle = F$. Clearly $f(u) = 0$, since $f(x) \in \langle f\rangle$. Since $a \mapsto a +\langle f\rangle$ for $a \in \Bbb Z_p$ is an embedding of $\Bbb Z_p$ in $F$, we can consider $F$ an extension of $\Bbb Z_p$.
Note that $\Bbb Z_p(u) \subseteq F$, and since $f$ is irreducible over $\Bbb Z_p$, we have: $[\Bbb Z_p(u): \Bbb Z_p] = \text{deg }f = m$. So, as a vector space over the field $\Bbb Z_p$, $\Bbb Z_p(u)$ has $m$ "coordinates" (we have $m$ basis vectors), and we may freely choose any of the $p$ elements of $\Bbb Z_p$ for each coordinate, giving $p^m$ vectors in all. On the other hand, elements of $F$ are of the form:
$a_0 + a_1x +\cdots + a_{m-1}x^{m-1} + \langle f\rangle$
$= a_0 + a_1u + \cdots + a_{m-1}u^{m-1}$, so we see the powers of $u$ span $F$, thus $\Bbb Z_p(u) = F$ ($\Bbb Z_p$-linear independence of these powers is guaranteed by the irreducibility of $f$, which is of minimal degree for $u$).
Now $F^{\ast}$ is a finite group (under multiplication) of order $p^m - 1$. It follows (from Lagrange) that for all $\alpha \in F^{\ast}$ we have:
$\alpha^{p^m - 1} - 1 = 0$, and thus all elements (including $u$) of $F$ are roots of:
$x(x^{p^m - 1} - 1) = x^{p^m} - x$.
Since $f$ is the minimal polynomial for $u$ (over $\Bbb Z_p$), it follows that $f|(x^{p^m} - x)$, that is, we may take $n = \text{deg }f$ (minimal polynomials divide any annihilating polynomial-this is easy to see from the division algorithm and the minimal degree of $f$).
