Property of irreducible polynomials in $Z_p[x]$ Let $p$ be a prime, and $Z_p = \mathbb Z / p \mathbb Z$ be the finite field with $p$ elements.
I want to prove that every irreducible polynomial in $Z_p[x]$ is a divisor of $x^{p^n}-x$ for some positive integer $n$, without using field extensions. How could I do that?
 A: I'll write $\mathbf F_p$ instead of your $Z_p$ and assume that it stands for $\mathbf Z/p\mathbf Z$. I'm not sure what you mean by "without using field extensions", because that power of $p$ has to come from somewhere, but I'll assume that you mean "without needing to know anything about extensions of $\mathbf F_p$".

Let $f(x)$ be your polynomial and $d$ its degree. Since $f(x)$ is irreducible, the ideal $(f)$ it generates in $\mathbf F_p[x]$ is maximal, so $K=\mathbf F_p[x]/(f)$ is a field. For any polynomial $g(x)\in\mathbf F_p[x]$, I'll write $\overline{g(x)}\in K$ for its image modulo $(f)$.
The field $K$ is also a $d$-dimensional vector space over $\mathbf F_p$ because $(\overline 1, \overline x, \overline x^2, \ldots, \overline x^{d-1})$ is a basis, so it is finite with $p^d$ elements. It follows that $K^*$ is a group of order $p^d-1$, thus $a^{p^d-1} - 1 = 0$ for any $a\in K^*$, and so $a^{p^d} - a = 0$ for any $a\in K$. In particular, for $a = \overline x$, we get
\begin{align}\tag{*}
\overline x^{p^d} - \overline x = 0.
\end{align}
Here's the part that's a bit field-extension-y. There is a homomorphism of fields $\varphi:\mathbf F_p\to K$ defined by first identifiying an element $a\in\mathbf F_p$ with the constant polynomial $a\in\mathbf F_p[x]$, and then reducing modulo $(f)$. This gives you a homomorphism of rings $\widetilde\varphi:\mathbf F_p[x]\to K[x]$ defined by
\begin{align*}
  \widetilde\varphi(a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0)
= \varphi(a_n)x^n + \varphi(a_{n-1})x^{n-1} + \cdots + \varphi(a_0).
\end{align*}
For any polynomial $g(x)\in\mathbf F_p[x]$, write $\widetilde g(x) = \widetilde\varphi(g)\in K[x]$; then it turns out that $\widetilde g(\overline x) = \overline{g(x)}$. Applying this to $g(x) = x^{p^d} - x$, what $(*)$ tells us is that $\widetilde g(\overline x) = 0$, and so $\overline{g(x)} = 0$; that is to say, $g(x) = x^{p^d} - x$ is a multiple of $f(x)$.
