Example of a field on which every irreducible polynomial has degree a power of $p$

Exercise A-47 in Milne's Fields and Galois Theory notes asks to prove that if $$p$$ is a prime number and $$F$$ is a field of characteristic zero such that every irreducible polynomial $$f(X)\in F[X]$$ has degree $$p^n$$ for some non negative integer $$n$$, then every polynomial equation with coefficients in $$F$$ is solvable by radicals.

I already solved the exercise. The key is to show that the Galois group $$\mathrm{Gal}(\overline{F}/F)$$ is a $$p$$-group, where $$\overline{F}$$ is the algebraic closure of $$F$$.

Question: For a fixed prime $$p$$, is there any field $$F$$ satisfying the given hypothesis?

For example, for $$p=2$$ the answer is yes: any real closed field satisfies the hypothesis. What are such examples for other primes?

Hint: You can try to look at the union of the fields $$\mathbb{F}_{l^n}$$ inside $$\overline{\mathbb{F}}_l$$, for $$n$$ powers of $$p$$.
$$\bf{Added:}$$ for a more sophisticated example, see Iwasawa theory.
$$\bf{Added:}$$ Jyrki corrected me, so the above stuff has only documentary value. Now I have an idea, not sure how good it is:
Consider a field $$F$$ and its algebraic closure $$\bar F$$. The Galois group $$G=\operatorname{Gal}(\bar F/F)$$ is a profinite group. I think for these kind of groups we have Sylow subgroups. So let $$P$$ a $$p$$-Sylow subgroup of $$G$$. The subfield $$K$$ of $$\bar F$$ invariant by $$P$$ has absolute Galois group $$\operatorname{Gal}(\bar F/K) = P$$. It follows that every finite Galois extension of $$K$$ has degree a power of $$p$$.
Obs: in fact we don't need $$P$$ to be a Sylow pro-$$p$$ group, just simply a pro-$$p$$ group ( I vaguely understand this). But $$K$$ will corresponding to a Sylow will be the smallest inside $$\bar F$$ (up to conjugation).
• But that is not a field of characteristic zero, which is one of the hypothesis imposed on &F$. Commented May 11, 2023 at 22:54 • @Albert: Oh, I missed that. But we are still winning, with unramified extensions of$\mathbb{Q}_l$. Commented May 12, 2023 at 0:51 • I think that in your example you want the union of fields$\Bbb{F}_{\ell^n}$,$n$coprime to$p$as opposed to powers of$p$. Commented May 12, 2023 at 3:59 • @Jyrki Lahtonen: Yes, I see, you are probably very right. My example gets every irreducible of degree prime to$p$. Maybe we can even do it for a subset$P$of primes ... Let me think a little bit...But now lifting to the p-adic field seems not so clear.. Commented May 12, 2023 at 4:05 • I agree that the$p\$-adic idea needs a different construction. Don't know what it would be :-( Commented May 12, 2023 at 4:12