(1) Let $L=K(\sqrt[n]{a})$ then $L/K$ is Galois since $K$ contains all $n^{\mathrm{th}}$ roots of unity and thus is a splitting field for $x^n-a$ which is separable since $n\neq 0$ in $K$ Thus the elements of $\Gal(L/K)$ permute the roots of $x^n-a$ which are $$\sqrt[n]{a},\ \zeta \sqrt[n]{a},\ \zeta^2 \sqrt[n]{a},\ \dotsc,\ \zeta^{n-1}\sqrt[n]{a}$$ and thus for $\sigma\in \Gal(L/K)$ we have $\sigma(\sqrt[n]{a}) = \zeta_{\sigma} \sqrt[n]{a}$ for some $\zeta_{\sigma}\in\boldsymbol{\mu}_n$ Define a map $$p:\Gal(L/K)\to \boldsymbol{\mu}_n:\sigma\mapsto\zeta_{\sigma}$$ We will show that $p$ is an injective homomorphism, which proves the result.

Since $\boldsymbol{\mu}_n\subset K$ each $n^{\mathrm{th}}$ root of unity is fixed by $\Gal(L/K)$ Then for $\sigma,\tau\in\Gal(L/K)$ $$ \zeta_{\sigma\tau}\sqrt[n]{a} =\sigma\tau(\sqrt[n]{a}) = \sigma(\zeta_{\tau}\sqrt[n]{a}) = \zeta_{\tau}(\sigma(\sqrt[n]{a})) = \zeta_{\sigma}\zeta_{\tau}\sqrt[n]{a} $$ so that $\zeta_{\sigma\tau} = \zeta_{\sigma}\zeta_{\tau}$ and $p$ is a homomorphism. The kernel of the map consists of all elements of $\Gal(L/K)$ which fix $\sqrt[n]{a}$ so that $p$ is injective and we are done.

(2) Note that $\N_{L/K}(\zeta) = 1$ since $\zeta$ is a root of $x^n-1$ so that by Hilbert's Theorem 90, $$\zeta = \sigma(u)/u,\quad\text{for some }u\in L$$ But then $\sigma(u) = \zeta u$ so that $\sigma(u^n) = \sigma(u)^n = \zeta^n u^n = u^n$ and $a=u^n\in K$ since it is fixed by a generator of $\Gal(L/K)$ Then clearly $K(u)$ is a splitting field of $x^n-a$ and the elements of $\Gal(L/K)$ send $u$ into distinct elements of $K(u)$ Thus $K(u)$ admits at least $n$ automorphisms over $K$ so that $[K(u):K]\geq n = [L:K]$ But $K(u)\subset L$ so $K(\sqrt[n]{a})=K(u)=L$