Let $K$ be a field of characteristic $p > 0$, and $f ∈ K[X]$ monic and irreducible with root $\alpha$. Let $F$ be the Frobenius endomorphism.
To demonstrate: $\alpha ∈ F[K(\alpha)] \Rightarrow f ∈ F[K][X]$.
Assume the antecedent. Then we know a number of things. The first is that there exists some $x ∈ K(\alpha)$ such that $\alpha = x^p$. Let $k := [K(\alpha):K]$. Then there exist $a_0, a_1, \ldots, a_k∈ K$ such that $$ \alpha = x^p = \left(a_0 + a_1·\alpha + \ldots + a_k· \alpha^k\right)^p = a_0^p + a_1^p·\alpha^p + \ldots + a_k^p ·\alpha^{pk}. $$
Second, as $f$ is monic and irreducible, it must be the minimal polynomial of $\alpha$ over $K$: $f = f^{\alpha}_K$ and $k = \deg f$. This yields, for $f = ∑_{i=0}^k f_iX^i$: $$0 = f(\alpha) = f_0 + f_1\alpha + \ldots +f_k\alpha^k $$
There are some more harmless facts:
III) If $f$ is inseparable then the consequent cannot be true: not all coefficients of $f$ are $p^{\text{th}}$ powers in $K$. Moreover, it is of the form $g(X^p)$ for some $g ∈ K[X]$.
IV) There are $n ∈ ℕ$ and separable irreducible $g ∈ K[X]$ such that $f = g\left(X^{p^n}\right)$.
V) There is an $m ∈ ℕ$ such that deg$(f) ∈ p^mℤ$ ($∵$ as $K(\alpha)/K$ is finite, $[K(\alpha):K] = [K(\alpha):K]_s · p^m$.)
My first idea was to write $$ a_0^p - \alpha + a_1^p·\alpha^p + \ldots + a_k^p ·\alpha^{pk} = x^p - \alpha = 0 = f(\alpha) = f_0 + f_1\alpha + \ldots +f_k\alpha^k $$ and try to collect like powers. However, this leads to all kinds of cases to distinguish (e.g. $k < p, k=p, k >p$) and it quickly becomes a mess.
I don't think that all the facts given here are needed. Neither do I think there are arguments needed that are not given or cannot be deduced from these, but I have a hard time organising these relationships and dependencies in this new world.
With $f\in K[x],g\in K^p[x]$ the $K$ and $K^p$ minimal polynomials of $a$ then $f | g$ in $K[x]$ and $f\in K^p[x]$ iff $f=g$ iff $$\deg(g)=[K^p(a):K^p]=[K(a):K]=\deg(f)$$
For any finite characteristic $p$ extension $L/M$ then $[L:M]=[L^p:M^p]$.
$K(a)^p=K^p(a^p)$