Note that you are not asked to prove that there is a chain of simple irreducible radical extensions $\Bbb Q \subset E_1 \subset E_2 \subset \ldots \subset \Bbb Q(\zeta_n)$, but that there is a chain $\Bbb Q \subset \ldots \subset L$ such that $\Bbb Q(\zeta_n) \subset L$.
If a normal extension $K \subset L$ has Galois Group isomorphic to $\Bbb Z/p \Bbb Z$, then $K(\zeta_p) \subset L(\zeta_p)$ is a normal simple irreducible radical extension :
$K \subset K(\zeta_p)$ is a normal extension, thus $Gal_{K(\zeta_p)}(L(\zeta_p))$ and $Gal_L(L(\zeta_p))$ are normal subgroups of $Gal_K(L(\zeta_p))$ and there is a well-defined restriction map $Gal_K(L(\zeta_p)) \to Gal_K(L) \times Gal_K(K(\zeta_p))$.
This map is injective because $L$ and $\zeta_p$ generate $L(\zeta_p)$. Furthermore, the composition of this map with either projection has to be surjective.
Since $|Gal_K(L)|=p $ and $|Gal_K(K(\zeta_p))|$ are coprime, the only subgroups of their product are products of subgroups, so the map has to be surjective. Thus it is an isomorphism.
let $x \in L \setminus K$. Then $K(x) = L$. Let $\sigma$ be a generator for $Gal_K(L) = Gal_{K(\zeta_p)}(L(\zeta_p))$, and look at $y = \sum_{k=0}^{p-1} \zeta_p^{-k} \sigma^k(x) \in L(\zeta_p)$. Then, $\sigma(y) = \sum_{k=0}^{p-1} \zeta_p^{-k} \sigma^{k+1}(x) = \zeta_p y$.
Thus $\sigma(y^p) = \sigma(y)^p = \zeta_p^p y^p = y^p$. Since $y^p$ is fixed by $\sigma$, $y^p \in K(\zeta_p)$.
Since $X^p - y^p = \prod(X - \zeta_p^k y) = \prod(X - \sigma^k(y))$, this polynomial must be irreducible over $K(\zeta_p)$ (there is exactly one orbit of the action $Gal_{K(\zeta_p)}(L(\zeta_p))$ on its roots)
Finally, $K(\zeta_p,y) = L(\zeta_p)$ because there is an obvious inclusion and their degree over $K(\zeta_p)$ are the same.
Next, if $K \subset L$ is a normal simple irreducible radical extension of degree $p$,
and $K \subset K'$ is a normal extension, then $K' \subset K'L$ is still a normal simple irreducible radical extension, of degree $1$ or $p$. The polynomial $X^p - y^p$ can't suddenly become reducible unless all the $y$ are in $K'$, because $\Bbb Z/p\Bbb Z$ has no nontrivial subgroup.
With this, you can show that if $K \subset L$ and $L \subset M$ are solvable by simple irreducible radical extensions, then so is $K \subset R$ (just add all the roots one after the other).
Finally, with an induction argument, you can finally show that since $K \subset K(\zeta_p)$ is abelian of degree dividing $p-1$, it is solvable, and since $K(\zeta_p) \subset L(\zeta_p)$ is solvable, $K \subset L(\zeta_p)$ is solvable, which implies that $K \subset L$ is too. And by decomposing any abelian extension into cyclic prime extensions, you obtain that any abelian extension is solvable.
Now we apply this to $\Bbb Q \subset \Bbb Q(\zeta_{47})$. Since its Galois group is isomorphic to $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 23 \Bbb Z$, we need to add $\zeta_2$ (which is already there, it's $-1$) and $\zeta_{23}$. For this one, we need $\zeta_{11}$, which needs $\zeta_5$, which needs $\zeta_4$.
The resulting chain (showing the degrees of the extensions) is :
$\Bbb Q \subset^2 \Bbb Q(\sqrt{-1}) \subset^2 \Bbb Q(\sqrt{-1},\sqrt 5) \subset^2 \Bbb Q(\zeta_{20}) \subset^2 \Bbb Q(\zeta_{20}, \sqrt{-11}) \subset^5 \Bbb Q(\zeta_{220}) \subset^2 \Bbb Q(\zeta_{220}, \sqrt{-23}) \\ \subset^{11} \Bbb Q(\zeta_{5060}) \subset^2 \Bbb Q(\zeta_{5060}, \sqrt{-47}) \subset^{23} \Bbb Q(\zeta_{237820}) \supset \Bbb Q(\zeta_{47})$.
Theoretically, you can compute explicitly at each step what are the things you are taking $n$th roots of and express everyone in terms of radicals, though it gets messy really fast.
(I only explicited the quadratic subfields of the $\mathbb Q(\zeta_p)$)
A very simple counting estimation will show that such polynomials have to exist. Let $q=p^k$ and $F=\Bbb F_q$, then it is known that $X^{q^n}-X$ is the product of all irreducible monic polynomials over$~F$ of some degree$~d$ dividing $n$. The product$~P$ of all irreducible monic polynomials over$~F$ of degree strictly dividing $n$ then certainly divides the product over all strict divisors$~d$ of$~n$ of $X^{q^d}-X$ (all irreducible factors of$~P$ are present in the latter product at least once), so that one can estimate
$$
\deg(P)\leq\sum_{d\mid n, d\neq n}\deg(X^{q^d}-X)\leq\sum_{i<n}q^i=\frac{q^n-1}{q-1}<q^n=\deg(X^{q^n}-X),
$$
so that $P\neq X^{q^n}-X$, and $X^{q^n}-X$ has some irreducible factors of degree$~n$.
I should add that by starting with all $q^n$ monic polynomials of degree $n$ and using the inclusion-exclusion principle to account recursively for the reducible ones among them, one can find the exact number of irreducible polynomials over $F$ of degree $n$ to be
$$
\frac1n\sum_{d\mid n}\mu(n/d)q^d,
$$
which is a positive number by essentially the above argument (since all values of the Möbius function $\mu$ lie in $\{-1,0,1\}$ and $\mu(1)=1$). A quick search on this site did turn up this formula here and here, but I did not stumble upon an elementary and general proof not using anything about finite fields, although I gave one here for the particular case $n=2$. I might well have overlooked such a proof though.
Best Answer
Let $E$ be the splitting field of $x^n-1\in F[x]$.
Let $p=char(f)$, if $p\ne 0$ then let $n=p^r s$ with $p\nmid s$, otherwise let $s=n$. Then $E=F(\zeta_s)$, $E/F$ is Galois, and $Gal(E/F)$ is a subgroup of $\Bbb{Z}/s\Bbb{Z}^\times$. Whence $[E:F]$ divides $\phi(s)$ which divides $\phi(n)$.
Given $u^n\in E$, let $f\in E[x]$ be the minimal polynomial of $u$, we have $f(0)= u^{\deg(f)} \zeta_s^l$ so that $u^{\deg(f)}\in E$ which implies that $u^{\gcd(n,\deg f)}\in E$. Therefore $[E(u):E]=\deg f$ divides $n$.
$[F(u):F]$ divides $[E(u):F]$ which divides $n \phi(n)$.
If $m$ doesn't divide $n \phi(n)$ then it can't be that $[F(u):F]=m$.