Kummer extension correspondence without roots of unity (Serge Lang)

I'm trying to solve the following problem.

Let $$k$$ be a field of characteristic $$0$$. Assume that for each finite extension $$E$$ of $$k$$, the index $$(E^* : E^{*n})$$ is finite for every positive integer n. Show that for each positive integer $$n$$, there exists only a finite number of abelian extensions of $$k$$ of degree $$n$$.

If $$k$$ contains a primitive n-th root of unity, one could use the one-to-one correspondence of abelian extension of $$k$$ of exponent n and subgroups of $$k^*$$ containing the n-th powers of the nonzero elements of $$k$$. For this case one of the ways to solve is as in the answer of this post: Find the bijection between Kummer's field and Galois subgroup.

But for $$k$$ not containing n-th roots of unity, do we have any kind of correspondence between, say, abelian extension of $$k$$ of exponent m and abelian extension of $$k(\zeta)$$ of exponent n, whence $$\zeta$$ is a primitive n-th root of unity?

I observed that an abelian extension of $$k$$ of exponent n has extension degree no more than the extension degree over $$k(\zeta)$$ of the abelian extension of $$k(\zeta)$$ of exponent n generated by the same set, multiplied by $$\varphi(n)$$, whence $$\varphi(n)$$ denotes the Euler function.

Another observation: Assume $$k$$ does not contain n-th roots of unity. Let H be a subgroup of $$k^*$$ containing the n-th powers of the nonzero elements of $$k$$, then $$H$$ and $$\zeta^j$$ together generates a subgroup of $$k(\zeta)^*$$ containing the n-th powers of the nonzero elements of $$k(\zeta)$$.

Let $$L/k$$ be the compositum of all the abelian extensions of degree at most $$n$$ over $$k(\zeta_n)$$. Since $$k$$ has characteristic zero, $$L/k$$ is separable. Then, since $$k(\zeta_n)$$ has all $$n$$-th roots of unity, you already know that $$L/k$$ is finite. If $$E/k$$ is an abelian extension of degree $$\leq n$$, then $$E(\zeta_n)$$ is an abelian extension of $$k(\zeta_n)$$ of degree $$\leq n$$, hence $$E\subset E(\zeta_n) \subset L$$. Since $$L/k$$ is separable, it contains at most finitely many subextensions. Hence the set of possible $$E$$ is finite.