I saw a rather sleek proof of the following fact:
Let $K$ be a number field. Let $L/K$ be a maximal abelian unramified outside $S$ extension of exponent $m$ of $K$ where $S$ is a finite set of places of $K$. Then $L/K$ is finite.
The above result is used, for example, in the proof of the weak Mordell-Weil theorem.
Here's the proof. Let me know if it looks good.
By GCFT we have surjection
$$ \frac{\mathbb I_K }{K^* K_{\infty +}^*\prod _{v\not \in S}\mathcal O_{K_v}\prod _{v\in S \backslash \infty } {K_v^*}^m}\rightarrow \operatorname{Gal}(L/K)$$
So it suffices to show that the LHS is finite. We have by strong approximation an injection
$$\frac{\mathbb I_K }{K^* K_{\infty +}^*\prod _{v\not \in S}\mathcal O_{K_v}\prod _{v\in S \backslash \infty } {K_v^*}^m} \hookrightarrow \frac{\prod _{v\in S \backslash \infty}K_v^*}{\prod _{v\in S \backslash \infty}{K_v^*}^m} $$ The RHS is finite since each $K_v^*/{K_v^*}^m$ is $\mu_m(K_v)$ -finite by kummer theory.
Edit The above inclusion is clearly wrong as pointed out by @Aphelli. However, this may work $$\frac{\mathbb I_K}{K^* \prod _v \mathcal O_{K_v}}$$ is finite since it is isomorphic to the ideal class group. There is a finite index quotient of $K^*\prod _v \mathcal O_{K_v}$, namely by the subgroup $K^* K_{\infty +}^* \prod _{v\not \in S} \mathcal O_{K_v}^* \prod _{v\in S \backslash \infty } U_n(K_v)$ call this $H$. Since the quotient in the surjection is smaller than the quotient by $H$ of $\mathbb I_K$, it must be finite.