Irreducible polynomial over field of order p Let $p$ be a prime and $F=\mathbb{Z}/p\mathbb{Z}$ and $f(t)\in F[t]$ be an irreducible polynomial of degree $d$.
I need to show that $f(t)$ divides $t^{(p^{n})}-t$ if and only if $d$ divides $n$.
 A: The first theorem of the following paper provides the proof:
http://www.jstor.org/stable/2316211
Or look at Theorem $7.6$ of the following article 
http://www-groups.mcs.st-and.ac.uk/~neunhoef/Teaching/ff/ffchap3.pdf
A: @pritam: Thanks for the resource!
For the users who might want to know the answer from the link, here it is:
Suppose $f$ divides $t^{(p^{n})}−t$. Then, if $a$ is a root of $f$, then $a^{(p^{n})}=a$, so $a$ is contained in a field of order $p^{n}$ so $F[a]$ is a subfield of $F^{'}=\mathbb{Z}/p^{n}\mathbb{Z}$. But since $f$ is a polynomial of degree $d$, $[F[a]:F]=d$ and $[F^{'}:F]=n$ and so $d$ divides $n$ since $n=[F^{'}:F[a]]d$.
Conversely, suppose $d$ divides $n$. Then $F^{'}=\mathbb{Z}/p^{n}\mathbb{Z}$ contains $F^{''}=\mathbb{Z}/p^{d}\mathbb{Z}$ as a subfield. If $a$ is a root of $f$, then $F[a]=F^{''}=\mathbb{Z}/p^{d}\mathbb{Z}$. Thus, $a\in F^{''}$ so $a\in F^{'}$. So $a^{(p^{n})}=a$ so every root of $f$ is a root of $t^{(p^{n})}-t$. Thus $f$ divides $t^{(p^{n})}-t$.
