I'm trying to understand the proof of this fact, that proceeds in the following way:
$K(\sqrt[n]{\Delta})\subseteq L$ because $\sqrt[n]{\Delta}\subseteq L^*$. For the converse, let's view $L$ as the composite $L=\prod_i L_i$ where $L_i$ are the finite subextensions of $L/K$. Then $L_i$ is abelian, finite, of exponent $n$, so $\text{Gal}(L_i/K)$ is a cartesian product of cyclic groups of order $d \mid n$, and so we can view $L_i$ as the composite $L_i=\prod_j F_{ij}$ of the cyclic extensions corresponding to the cyclic subgroups of $\text{Gal}(L_i/K)$. Then, because of the Kummer theorem, every cyclic extension $F_{ij}$ is $F_{ij}=K(\sqrt[n]{c})$ for some $c \in K^*$. So $L$, as a composite of such extensions, is contained in $K(\sqrt[n]{\Delta})$.
The problem is in the use of the Kummer theorem. Why we have that every $F_{ij}$ is of the type $K(\sqrt[n]{c})$? This seems to occur only if $[F_{ij}:K]=n$, otherwise the degree of the extension is just a divisor $d$ of $n$, and $K(\sqrt[d]{c})\subseteq K(\sqrt[n]{c})$, so where is the problem?
Just for clarity, this is the statement of the Kummer theorem which I'm referring to:
Let $K$ be a field and $n$ a positive integer such that $\text{char } K \nmid n$ and $\zeta_n \in K$, where $\zeta_n$ is a primitive $n$th rooth of unity. If $L/K$ is a cyclic extension with $[L:K]=n$, then $L=K(\sqrt[n]{c})$ for some $c \in K$.
Obviously in the hypothesis we assume that $\text{char } K \nmid n$ and $\zeta_n \in K$.
The Kummer theorem gives $x\in K^*$ such that $F_{ij}=K(\sqrt[d]{x})$. Now note that since $d$ divides $n$, we can write $d=mn$ and then a $d$th root of $x$ is also an $n$th root of $x^m$, so you can also describe $F_{ij}$ as $K(\sqrt[n]{c})$ where $c=x^m$.