I'm in the middle of the proof about $P_n = x^{p^n} - x \in F_p[x]$ for p prime, being the product of all monic, irreducible polynomials of degree $d$ such that $d \mid n$.
I've already proved that if $\pi(x) \in F_p[x]$ is monic, irreducible, such that $\deg(\pi(x)) \mid n$ then $\pi(x) \mid P_n$.
I've also proved that there are no irreducible factors with multiplicity in the factorization of $P_n$.
It remains to prove that every factor of $P_n$ is of the form $\pi(x)$, that is, if $r(x) \in F_p[x]$ is a monic, irreducible polynomial of degree $d$, such that $r(x) \mid P_n$, than $d | n$.
How can I go on with that to finish the proof? Here's what I was trying:
Let $\pi(x) \in F_p[x]$ be a monic and irreducible polynomial of $\deg(\pi(x)) = d$, such that $\pi(x) | P_n$. Since $\pi(x) | P_d$, it implies that $\pi(x) \mid P_{\gcd(n,d)}$. Now suppose that $d \nmid n$, therefore $\gcd(n,d) < d$...
(Now I'm aiming for a contradiction in the fact that $\pi(x) \mid P_{\gcd(n,d)}$ or $\pi(x) \mid P_n$, but idk how to continue...)
I'm new to finite fields and abstract algebra, so please, if it's possible, don't use more advanced techniques.
Thanks for your time and help.
Consider the following field : $F:=\mathbb F_p[x]/(r)$ (it's a field because $r$ is irreducible of course).
This field has $p^d$ elements, therefore by Lagrange's theorem applied to $F^\times$, we see that each of its elements is a root of $X^{p^d}-X$, so $X^{p^d}-X = \prod_{a\in F}(X-a)$
Now note the following thing : in a characteristic $p$ field, $(a+b)^{p^k} = a^{p^k}+ b^{p^k}$ for any integer $k$.
It follows that if we let $\alpha$ denote the image of $x$ in $F$, and if $P$ is a polynomial, then $P(\alpha)^{p^n} = P(\alpha^{p^n}) = P(\alpha)$ (indeed, $\alpha^{p^n} = \alpha$ because $r\mid P_n$, so $\alpha$ is a root of $p^n$).
Therefore, $P(\alpha)$ is also a root of $P_n$. But $F$ consists of elements of this form : any element of $F$ is a root of $P_n$ !
Therefore $P_d\mid P_n$, and it's easy to see that this divisibility relation holds in $\mathbb F_p[x]$, not only $F[x]$ (to see why, think about euclidean division).
Do you know that this implies $d\mid n$ ? (I think you do, since you mentioned $P_{gcd(d,n)}$) If yes, then we are done. Else, I'll add something to that effect.