Kummer extensions I want to prove the following:

If $F$ contains all $n$-roots of unity and $\operatorname{char}F$ not divides $n$ then $K:=F(\sqrt[n]{a_1},\sqrt[n]{a_2},\ldots,\sqrt[n]{a_m})/F$ is a Galois abelian extension.

Hint: Prove that $G:=Gal(K/F)\cong \left<a_1 F^{\times n},\ldots,a_m F^{\times n}\right>\leq  F^\times/ F^{\times n}$, where $ F^{\times n}=\{x^n: x\in F^\times\}$.
Attempt:
I show that $K/F$ is galois,  $F^{\times n}$ subgroup, but I tried (and I can't) to prove hint for $m=1$ considering the map $\psi\colon \left<a F^{\times n}\right>\to G=F(\sqrt[n]{a})/F$ given by $a^kF^{\times n}\to\sigma_k$ where $\sigma_k$ is defined by $\sigma_k(\sqrt[n]{a})=\zeta_n^{mk}\sqrt[n]{a}$ where $m$ is the order of $aF^{\times n}$. I proved that $\psi$ is an homomorphism but I don't know how to prove that such $\sigma_k$ is an element of $G$, I tried using the basic theorems of radical extensions. I am almost sure that such a map is the desired isomorphism since I proved with a lot of examples.
 A: Definition. Let $L$ be a field and let $\alpha$ be an element of
a field extending $L$. The radical degree of $\alpha$ over $L$
is the smallest $p\geq 1$ such that $\alpha^p\in L$ (or $+\infty$ if
no such $p$ exists).
Then we have:
Lemma. If the radical degree $d$ of $\alpha$ is finite and not divisible
by the characteristic of $L$, and if $L$ contains all
$d$-roots of unity, then the minimal polynomial of
$\alpha$ over $L$ is exactly $X^d-{\alpha}^d$. In particular, the
radical degree coincides with the usual degree.
Proof of lemma. The minimal polynomial $M$ of
$\alpha$ over $L$ divides $X^d-{\alpha}^d$, so it is the form
$M=\displaystyle\prod_{k\in I}(X-\zeta^k\alpha)$ where $\zeta$ is 
a primitive $d$-th root of unity and $I$ is a subset of
$\lbrace 1,2, \ldots ,d \rbrace$. Expanding, we see that
$$
M=X^{|I|}+\sum_{j=0}^{|I|-1}Q_k(\zeta)\alpha^{|I|-j}X^j
$$ 
where the $Q_k$ are polynomials with integer coefficients. Now the
coefficient $Q_k(\zeta)\alpha^{|I|-j}$ can be in $L$ only
if it is zero or if $|I|-j=d$. There is at least one non-zero coefficient
besides the leading one, and this can happen only when $|I|=d$ and $j=0$, which 
concludes the proof.
Note that $\lbrace p\in {\mathbb Z} \mid \alpha^p \in L\rbrace$
is a subgroup of $\mathbb Z$, so it can only be $d{\mathbb Z}$. It follows that :
Corollary. The lemma still holds if the hypothesis on $\alpha$ is weakened
to $\alpha^n\in L$ for some $n>0$ not a multiple of the characteristic of $L$.
Now we can iterate the corollary: if we denote by $d_k (1\leq k\leq m) $ the radical degree of $a_k^{\frac{1}{n}}$ over $F_k=F(a_1^{\frac{1}{n}}, \ldots ,a_{k-1}^{\frac{1}{n}})$,
then by the corollary $d_k$ is in fact the degree of $a_k^{\frac{1}{n}}$ over $F_k$.
We have $[K:F]=[F_{m+1}:F_1]=\displaystyle\prod_{j=1}^{m}[F_{j+1}:F_j]=d_1d_2\ldots d_m$.
So the Galois group $G={\sf Gal}(K/F)$ contains exactly $d_1d_2\ldots d_m$ elements. Let 
$H=\frac{\mathbb Z}{d_1\mathbb Z} \times \ldots
\times \frac{\mathbb Z}{d_m\mathbb Z}$, and let $\zeta$ is a fixed primitive $n$-th root of unity. Let $h=(h_1,h_2,\ldots ,h_m)\in H$.
The minimal polynomial of $\alpha_1$ over $F_0=F$ is exactly $X^{d_1}-\alpha_1^{d_1}$, so
there is a unique $i_1\in {\sf Gal}(F_1/F)$ satisfying $i_1(\alpha_1)=\zeta^{\frac{nh_1}{d_1}}\alpha_1$.
The minimal polynomial of $\alpha_2$ over $F_1$ is exactly $X^{d_2}-\alpha_2^{d_2}$, so
there is a unique $i_2\in {\sf Gal}(F_2/F)$ extending $i_1$ and satisfying $i_2(\alpha_2)=\zeta^{\frac{nh_2}{d_2}}\alpha_2$.
Iterating this process, we see eventually that there is a unique 
$\sigma\in {\sf Gal}(F_k/F)$ such that $\sigma(\alpha_k)=\zeta^{\frac{nh_k}{d_k}}\alpha_k$ for every $k$ between
$1$ and $m$. This defines an injective map $H \to G$. It will be bijective
because $|G|=|H|=d_1d_2\ldots d_m$, and it is very easy to check that it is 
a homomorphism.
So now it remains to construct an isomorphism between $H$ and
$W=\langle a_1F^{\times  n},a_2F^{\times  n},\ldots ,a_mF^{\times  m}\rangle$. It is easy to check that the map
$$
H \to W, \ (i_1,i_2,\ldots, i_m) \mapsto \langle a_1^{i_1}a_2^{i_2} \ldots a_m^{i_m}\rangle 
$$
is well defined and does the job.
