Factorization of cyclotomic polynomials Is the following statement correct?
Let $F$ be field with characteristic not equal to $p$ which does not contain any primitive $p$th root of unity.
Let  $\phi_{p}(X) = f_{1}(X)f_{2}(X)\cdots f_{k}(X)$ be the factorization of $p$th cyclotomic polynomial into irreducible factors over $F$. Then $\phi_{p^2(X)} = f_{1}(X^p)f_{2}(X^p)\cdots f_{k}(X^p)$ is the factorization of $p^2$th cyclotomic polynomial into irreducible factors over $F$.
If the statement is correct, then what is the idea of the proof?
 A: I will just elaborate on Felipe Voloch's and znt's comments.
Since $\phi_{p^2}(x) = \phi_p(x^p)$, we have the following: If $\prod f_i(x)$ is a factorization of $\phi_p(x)$ then indeed $\prod f_i(x^p)$ is a (not necessarily "final"!) factorization of $\phi_{p^2}(x)$. Hence, the question may be reformulated equivalently as follows: If $f(x)$ is the minimal polynomial of some root of $\phi_p(x)$, is $f(x^p)$ still irreducible?

(Classical) Lemma: Let $F=\mathbb{F}_q$, and let $n$ be coprime to $\text{char}(F)$. Then $\phi_{n}$ factors into $\frac{\phi(n)}{\text{ord}_n(q)}$ irreducible factor of degree $\text{ord}_n(q)$.  ($\text{ord}_n(q)$ denotes the multiplicative order of $q$ modulo $n$.)
Proof: To understand the factorization of $\phi_n$ in $F$, we need to recall the following: If $\alpha \in \overline{F}$, then the degree of the minimal polynomial of $\alpha$ over $F$ (which is immediately irreducible over $F$) is the number of elements in the orbit of $\alpha$ under the action of the Frobenius $x\mapsto x^{|F|}$. Let $\alpha$ be a root of $\phi_n$ in $\overline{F}$. The orbit of $\alpha$ is of size $d$ iff $\alpha^{q^d}=\alpha$ and $\alpha^{q^{d'}} \neq \alpha$ for $d' \mid d$. In other words, if $\alpha$ is a root of unity of order dividing $q^d-1$ but not $q^{d'}-1$ for $d' \mid d$. By definition, $\alpha$ is a root of unity of order $n$. So we are looking for the minimal $d$ such that $n \mid q^d-1$. This $d$ is the (multiplicative) order of $q$ modulo $n$. Since $\phi_n$ factors into a product of some minimal polynomials, all of which must be of degree $\text{ord}_n(q)$, we are done. $\blacksquare$

It follows from the lemma that each $f_i(x)$ is of degree $\text{ord}_{p}(|F|)$, which implies that $\deg f_i(x^p) = p \cdot \text{ord}_{p}(|F|)$. The lemma also implies that the irreducible factors of $\phi_{p^2}$ are of degree $\text{ord}_{p^2}(|F|)$. So you ask: do we always have $\text{ord}_{p^2}(|F|) =  p \cdot \text{ord}_{p}(|F|)$? Well, since $A \equiv 1 \bmod p \implies A^p \equiv 1 \bmod {p^2}$, it follows that $$\text{ord}_{p}(|F|) \mid \text{ord}_{p^2}(|F|) \mid p \cdot \text{ord}_{p}(|F|),$$
i.e. $$\text{ord}_{p^2}(|F|) \in \{ p \cdot \text{ord}_{p}(|F|), \text{ord}_{p}(|F|) \}.$$
Both options are possible! If, by chance, 
$$(*) |F|^{ \text{ord}_{p}(|F|) } \equiv 1 \bmod {p^2},$$
then $\text{ord}_{p^2}(|F|) = \text{ord}_{p}(|F|)$. This means that each $f_i(X^p)$ is reducible (factors into $p$ factors of degree $\frac{\phi(n)}{\text{ord}_{p}(|F|)}$). Since $\text{ord}_{p}(|F|) \mid p-1$ and $(p-1,p)=1$, it follows that $(*)$ is equivalent to 
$$|F|^{p-1} \equiv 1 \bmod {p^2}.$$
In other words, the only counterexamples to your question are primes $p$ which are "Wieferich primes in base $|F|$". It's a conjecture that for every natural number $a$, there are infinitely many Wieferich primes in base $a$. See this Wikipedia article.
