TITLE: Showing that $\mathbb{Q}(\zeta_p, \sqrt[p]{\ell}) = \mathbb{Q}(\zeta_p + \sqrt[p]{\ell})$ for $p,\ell$ primes.
QUESTION [8 upvotes]: We consider the polynomial $x^p - \ell$, where $p,\ell$ are both prime numbers. Let $\zeta_p$ be a $p$-th root of unity. We wish to show that $L = \mathbb{Q}(\zeta_p, \sqrt[p]{\ell})$ is the same as $K = \mathbb{Q}(\zeta_p + \sqrt[p]{\ell})$. Here are some of the facts I have already proved:

The polymial $x^p - \ell$ is irreducible over $\mathbb{Q}$.
$[\mathbb{Q}(\zeta_p) : \mathbb{Q}] = p-1$ and $[\mathbb{Q}(\sqrt[p]{\ell}):\mathbb{Q}] = p$.
$L = \mathbb{Q}(\zeta_p,\sqrt[p]{\ell})$ is the splitting field of $x^p - \ell$ and has degree $[\mathbb{Q}(\zeta_p,\sqrt[p]{\ell}) : \mathbb{Q}] = p(p-1) = p^2 - p$. Furthermore, $L$ is a Galois extension of $\mathbb{Q}$.

My first attempt was to show both inclusions. One is trivial, but the other one is much, much harder, i.e. showing $\mathbb{Q}(\zeta_p,\sqrt[p]{\ell}) \subseteq \mathbb{Q}(\zeta_p + \sqrt[p]{\ell})$. I thought expanding $(\zeta_p + \sqrt[p]{\ell})^p$ with the binomial expansion would be useful, but it turns out to be uglier than I expected.
I further notice that one way to reduce the problem is to show that the only automorphism $\sigma \in \mathrm{Aut}(L/\mathbb{Q})$ that fixes $\zeta_p + \sqrt[p]{\ell}$ is the identity map $1$, so that $L$ and $K$ must coincide. Since $\sigma(\zeta_p + \sqrt[p]{\ell}) = \sigma(\zeta_p) + \sigma(\sqrt[p]{\ell})$ and $\sigma$ only permutes the roots, we see  that $\zeta_p + \sqrt[p]{\ell} \mapsto \zeta_p^m + \zeta_p^n\sqrt[p]{\ell}$ under $\sigma$ for some $m,n \in \{1,\ldots,p-1\}$. Hence, it suffices to show that
$$\zeta_p + \sqrt[p]{\ell} = \zeta_p^m + \zeta_p^n\sqrt[p]{\ell} \iff (m,n) = (1,0)$$
to conclude. However, I don't know where to go from there.
Any help or hint would be greatly appreciated. If there is an easier to show this, please let me know.

REPLY [9 votes]: You were very near to the full solution. If $n\neq 0$, you can rewrite your equality as
$$
\sqrt[p]{\ell}=\frac{\zeta_p-\zeta_p^m}{1-\zeta_p^n}
$$
So you deduce $\sqrt[p]{\ell} \in {\mathbb Q}(\zeta_p)$, and it is very easy to finish from there.