Show that $x^{p^n}-x$ is the product of all monic irreducible polynomials in $\mathbb{Z}/p\mathbb{Z}[x]$ of a degree $d$ dividing $n$. [duplicate]

I know that if $F$ is a field of $p^n$ elements contained in an algebraic closure $\overline{\mathbb{Z}/p\mathbb{Z}}$ of $\mathbb{Z}/\mathbb{Z}p$, then the elements of $F$ are the zeros $x^{p^n}-x$, and also that the degree of $F$ over $\mathbb{Z}/p\mathbb{Z}$ is $n$, but I don't know where to go from here. Since the elements of $F$ are the zeros of $x^{p^n}-x$ in $\mathbb{Z}/p\mathbb{Z}[x]$, that means $x^{p^n}-x$ is the product of polynomials whose zeros are the elements of $F$, but how do I know that the degree divides $n$?

• You are on the right track. If $x^{p^n}-x$ has an irreducible factor of degree $m$, then a zero of that factor generates a subfield $K$ with $p^m$ elements. Because $F$ is an extension field of $K$ we must have $|F|=|K|^\ell$ for some integer $\ell$. Therefore we must have $m\mid n$. Alternatively you can use the multiplicativity of extension degree in a tower of fields to reach the same conclusion. – Jyrki Lahtonen Feb 17 '16 at 20:29
• But, this question has been discussed many times on the site. I trust the other users found a good earlier incarnation. – Jyrki Lahtonen Feb 17 '16 at 20:31