If $\widetilde{L} =$ splitting field of all irreducible polynomials over $L$ of prime-power degree, is $\widetilde{\Bbb{Q}} = \overline{\Bbb{Q}}$? For a field $L$, let $\widetilde L$ be the splitting field of all irreducible polynomials over $L$ having prime-power degree.

Question: Do we have $\widetilde{\mathbf Q}=\overline{\mathbf Q}$?

My money is on "no", because I see no obvious reason why it should be true. If the answer is indeed negative, can one say what degree occurs as the smallest degree of an $f\in \mathbf Q[X]$ which does not split in $M$?
In any case, it seems quite difficult.
A variation:  We have a chain $$L \subset \widetilde L \subset \widetilde {\widetilde L} \subset \dots$$
Let $\widehat L$ be the limit of this chain. Is it even true that $\widehat{\mathbf Q}=\overline{\mathbf Q}$? Does the chain stabilize? The field $\widehat{\mathbf Q}$ has the strange property of having no finite extensions of prime-power degree. Correspondingly, the Galois group $\text{Gal }(\overline{\mathbf Q}/\widehat{\mathbf Q})$ has the strange property of having no open subgroups of prime-power index... 
For $L$ a finite field, it is easy to see that $\widetilde{L}=\overline{L}$. We obviously have $\widetilde{\mathbf R}=\overline{\mathbf R}$. I do not know if $\widetilde{\mathbf Q_p}=\overline{\mathbf Q_p}$, or if $\widehat{\mathbf Q_p}=\overline{\mathbf Q_p}$. (Edit: Every finite Galois extension of $\mathbf Q_p$ is solvable, and I believe it follows from this that $\widehat{\mathbf Q_p}=\overline{\mathbf Q_p}$.)
Addendum: The question has been answered (positively) by David Speyer at MO. (David: if would like to copy & paste your answer here as well, I will gladly accept it!)
 A: This is not a solution.
Here is what I hope is a correct restatement of the problem. Let me introduce the following conditions on an algebraic extension $L$ of $\mathbb{Q}$ which I don't have any imaginative names for:

P1: $L$ is the compositum of extensions of $\mathbb{Q}$ of prime power degree.
P2: $L$ fits into a tower of P1 extensions. Equivalently, $L$ is the compositum of towers of extensions of $\mathbb{Q}$ of prime power degree.

When a P1 (resp. P2) extension is Galois, its Galois group satisfies the following corresponding conditions:

P1: $G$ has subgroups $H_1, ..., H_n$ of prime power index with trivial intersection.
P2: $G$ has subgroups $H_1, ..., H_n$ with trivial intersection each of which fits into a tower $H_k \subseteq H_{k, 1} \subseteq H_{k, 2} \subseteq ... \subseteq H_{k, m} \subseteq G$ of subgroups each of which is of prime power index in the next.

The first question is equivalent to the question of whether every finite Galois extension of $\mathbb{Q}$ occurs as a subextension of a P1 extension. The second question is equivalent to the question of whether every finite Galois extension of $\mathbb{Q}$ occurs as a subextension of a P2 extension. These translate (imperfectly) into the following group-theoretic questions via the Galois correspondence:

Question: Is every finite group a quotient of a finite P1 (resp. P2) group?

(I say imperfectly because we also need both groups to occur compatibly as the Galois groups of a pair of finite Galois extensions of $\mathbb{Q}$. If the answer to the above question is "yes," we don't necessarily know that the answer to the original question is "yes," but if the answer to the above question is "no," we may be able to construct a counterexample to the original question conditional on solving a special case of the inverse Galois problem.)
Instead of solving the problem let me explain my mistake. I thought that "subextension of a P1 extension" was equivalent to "a P1 extension," but the obvious argument fails: if $L \subseteq \bigcup_i K_i$ (where by $\bigcup$ I mean compositum in a fixed Galois extension containing $L$ and the $K_i$) it does not follow that $L \subseteq \bigcup_i (L \cap K_i)$; an explicit counterexample is that
$$\mathbb{Q}(i) \subseteq \mathbb{Q}(\sqrt{2}) \cup \mathbb{Q}(\sqrt{-2})$$
but the corresponding compositum of intersections is just $\mathbb{Q}$.
The corresponding mistake on the finite group side is perhaps even more insidious: it is to incorrectly assume that every quotient of a P1 group is P1. One reason this is an attractive assumption is that every nilpotent group is P1 (since a finite nilpotent group is the direct product of its Sylow subgroups), and nilpotent is a property that passes to quotients!
Nevertheless the obvious argument fails: if $G \to G/N$ is a quotient map and $H_i$ are subgroups of $G$ of prime power index with trivial intersection, it's still true that their images $H_i N / N$ have prime power index in $G/N$ (this corresponds to taking the intersections $L \cap K_i$ in the above argument) but false that their intersection is necessarily trivial (this corresponds to the compositum failing to contain $L$ in the above argument) because $N$ can introduce additional relations. The example Galois corresponding to the above example is to take
$$G = C_2 \times C_2, H_1 = (0, 1), H_2 = (1, 0), N = (1, 1).$$
However, for all I know it might still be true that every quotient of a P1 (resp. P2) group is P1 (resp. P2). If someone has a proof of that it would imply that my original counterexample (the group $G = A_6$ is neither P1 nor P2 and occurs as the Galois group of an irreducible polynomial of degree $6$) still works. So:

Auxiliary question: Is every quotient of a finite P1 (resp. P2) group P1 (resp. P2)?

Note that a counterexample to the auxiliary question for P1 is necessarily not nilpotent. It's also the case that every solvable group is P2 and that this is also a property that passes to quotients, so a counterexample to the auxiliary question for P2 is necessarily not solvable.
