If $K(\zeta,\beta)/K$ is the splitting field for $f$, what can we say about $K(\zeta,\beta)/K(\zeta)$? Problem setup:

Let $n\in\mathbb{N}$ and let $K$ be a field whose characteristic does not divide $n$. The splitting field of $f=x^n-c$ ($c\not=0$) over $K$ is $K(\zeta,\beta)$ where $\zeta$ is a primitive $n$-th root of unity. Put $E=K(\zeta,\beta)$. What can we say about $[E:K(\zeta)]$?

Overall goal: I am working on a problem (with the same assumptions) where we assume $n=2^k$, and show $[E:K]$ is a power of $2$. We take the following steps:

Step 1: Assume $\zeta\in K$ and show that all irreducible factors of $f$ have degree $[E:K]$. Conclude that $[E:K]$ divides $n$.

I was able to do this fairly easily: first let $g$ be an irreducible factor of $f$, then let $\beta$ be a root of $g$. Since $\zeta\in K$, we have $E=K(\beta)$. Since $g$ is irreducible, $[E:K]=[K(\beta):K]=\deg(g)$. Since the degree of $f$ is the sum of the degrees of its irreducible factors, we know $[E:K]$ divides $n$.

Step 2: Assume $\zeta\not\in K$, suppose $n=2^k$, and show $[K(\zeta):K]$ is a power of $2$.

There is a more general result that says $[K(\zeta):K]$ divides $\phi(n)$ where $\phi$ is Euler's totient function (see here). Applying this, we obtain $[K(\zeta):K]$ divides $\phi(n)=\phi(2^k)=2^{k-1}$, so $[K(\zeta):K]$ is a power of $2$.
Where I get stuck/main question:

Final step: Apply steps (1) and (2) to show that if $n=2^k$, then $[E:K]$ is a power of $2$.

From step (1), the result follows quickly if $\zeta\in K$, since $[E:K]$ will divide $n$, a power of $2$, and hence $[E:K]$ will also be a power of $2$. I am getting confused about the case when $\zeta\not\in K$. If this is the case, then step (2) says $[K(\zeta):K]=2^\ell$ for some $\ell\leq k-1$. From the Degree Formula for Field Extensions, we have (let $\beta$ be a root of $f$):
$$
[E:K]=[K(\zeta,\beta):K]=[K(\zeta,\beta):K(\zeta)][K(\zeta):K]=[K(\zeta,\beta):K(\zeta)]2^\ell.
$$
If $[K(\zeta,\beta):K(\zeta)]$ is a power of $2$, we will be done. This leads me to my main question of this whole post, what can we say about $d=[K(\zeta,\beta):K(\zeta)]$? I know that if $f$ were irreducible over $K(\zeta)$, then $d=n=2^k$ and we're done. But we don't know that $f$ is irreducible over $K(\zeta)$. My idea is, maybe we can do something similar to Step (1) as follows:

Claim: Regard $f$ as a polynomial over $K(\zeta)$. Let $g$ be an irreducible factor of $f$ over $K(\zeta)$. Let $\beta$ be a root of $g$. Then $[K(\zeta,\beta):K(\zeta)]=\deg(g)$, so (using a similar argument as in Step (1)), $[K(\zeta,\beta):K(\zeta)]$ must divide $n=2^k$, and hence will also be a power of $2$.

Is this claim true? Or am I missing something here? If it is true, then the result follows immediately. It feels like my claim might be a slightly special case of a more general result, but I am not sure. I'd appreciate any feedback.
 A: There seems to be a mistake in Step 1 - I don't see how it follows that the degree of $g$ divides $n$ from what you've written.
In any case, you can use Galois theory to simplify both of your steps. In particular, $K(\zeta,\beta)/K(\zeta)$ is Galois and $K(\zeta)/K$ is Galois and it is easy to study these Galois groups.
Working with the first extension, we see that it is the splitting field of $x^n - c$ over $K(\zeta)$. In particular, if $\beta$ is some root of it, all the other roots are of the form $\zeta^i \beta$ and it splits after adjoining just $\beta$; the latter also means that the Galois group is determined by its action on $\beta$. If we take $\sigma$ in the Galois group, we see that it determines an element of $\mathbb Z/n\mathbb Z$ by $\sigma(\beta) = \zeta^i\beta$. One can check that this induces an injection of the Galois group into $\mathbb Z/n\mathbb Z$. Take $\sigma,\tau$ corresponding to $i,j$ and then:
$$\sigma\tau(\beta) = \sigma(\zeta^j \beta) = \zeta^j \sigma(\beta) = \sigma^{i+j}\beta$$
(note $\sigma$ fixes $\zeta$)
So the Galois group has order dividing $n$ by lagrange, and in this case it divides $2^k$ and hence has order some power of 2.
We can treat the latter extension in almost the same way. Again in this case the extension is obtained by adjoining $1$ primitive $n$th root of unity. Its conjugates are all of the form $\zeta^i$ for some $i$ coprime to $n$, and this gives rise to a natural embedding of the Galois group into $(\mathbb Z/n\mathbb Z)^\times $ by $\sigma\mapsto i$ when $\sigma(\zeta) = \zeta^i$. Just like above one can easily verify this is a homomorphism. Thus the Galois group has order dividing that of $(\mathbb Z/n\mathbb Z)^*$; the latter has order $\phi(n)$ which in this case is $2^{k-1}$ so again we get an extension of degree a power of $2$.
Finally, degrees multiply in towers and so the whole thing is degree a power of $2$.
