Irreducible factors of polynomial $X^{p^n}-X$ over $\mathbb{F}_{p^i}[X]$ I know how to compute the irreducible factor of polynomial $X^{p^n}-X$ where $p$ is prime over $\mathbb{F}_p[X]$. There is even a formula to know how many factors of each degree there are. 
How can I study the irreducible factors of polynomial $X^{p^n}-X$ over $\mathbb{F}_{p^i}[X]$ with $i> 0$
An example
Say I want to compute the irreducible factorization of $X^{16}-X$ over $\mathbb{F}_4$. I already know that over $\mathbb{F}_2$ this is $X(X-1)(X^2+X+1)(X^4+X^3+1)(X^4+X+1)(X^4+X^3+X^2+X+1)$. 
This other question claims that the factorization over $\mathbb{F}_4$ contains only factors of degree 1 or 2. Why is it the case?
 A: I have managed to proof that if I have an irreducible polynomial over $\mathbb{F}_{p}$ then over $\mathbb{F}^m$ its irreducible factors (if any) must all have same degree. The proof follows this answer. Basically I write $f = \prod q_i$ and take $\alpha$ a root of $q_1$, $\beta$ a root of $q_i$. Then since $f$ is irreducible we know that $Gal(f/\mathbb{F}_{p})$ must be a transitive subgroup and therefore there exists $\sigma$ in this group such that $\sigma(\alpha) = \beta)$. 
Then the extended $\sigma$ to polynomials gives $\overline{\sigma}(q_1) = h$ such that $h$ is irreducible, $\beta$ is a root of $h$ and $deg(h) = deg(q_1)$. Therefore $q_i | h$ and $deg(q_i) \le deg(q_1)$. Changing the roles of the elements implied should give the other inequality. 
For showing that they the same degree one proceeds as follows. We take a root of $f$, $\alpha$. Then $$n = [\mathbb{F}_p(\alpha):\mathbb{F}_p] = [\mathbb{F}_p(\alpha):\mathbb{F}_{p^m} \cap \mathbb{F}_p(\alpha)][\mathbb{F}_{p^m} \cap \mathbb{F}_p(\alpha):\mathbb{F}_p]$$
Now one uses what sometimes is called the Natural Irrationalities Theorem. Basically we want $[\mathbb{F}_p(\alpha):\mathbb{F}_{p^m} \cap \mathbb{F}_p(\alpha)] = [\mathbb{F}_{p^m}(\alpha):\mathbb{F}_{p^m}]$. 
In general, if $L/F$ is Galois and $K/F$ finite then $Gal(KL/K) \cong Gal(L/K \cap L)$. Above we just take $K = \mathbb{F}_{p^m}$ and $L = \mathbb{F}_p(\alpha)$.
Therefore, $$n = [\mathbb{F}_{p^m}(\alpha):\mathbb{F}_{p^m}][\mathbb{F}_{p^m} \cap \mathbb{F}_p(\alpha):\mathbb{F}_p]$$ Where the first part is precisely the degree of the irreducible polynomials $q_i$ therefore, now $[\mathbb{F}_{p^m} \cap \mathbb{F}_p(\alpha):\mathbb{F}_p]$ is precisely the degree of the irreducible polynomials of the extensions. We can take $\mathbb{F}_p(\alpha) \cong \mathbb{F}_{p^d}$ and therefore the intersection is $\mathbb{F}_{p^{gcd(d,m)}}$ and therefore $n = gcd(m,d)d$.
In my case, this leads to $4 = gcd(2,d)d$. Since $gcd(2,d) = 1,2$, possible solutions should verify $4 = 2d$ or $4 = d$. however $4$ is not a solution. Therefore, it has to be $d = 2$.
I wonder if the fact that finite fields are perfect fields can be exploited in some way...
For instance, I think that one could use that the Galois group of the extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is generated by the Frobenius automorphism $\phi$ to proof that all factors have the same number of zeros...
