Let $K=\mathbb{Q}(\sqrt[n]{a})$, where $a\in\mathbb{Q}$, $a>0$ and suppose $[K:\mathbb{Q}]=n$. Let $E$ be any subfield of $K$ and let $[E:\mathbb{Q}]=d$. Prove that $E=\mathbb{Q}(\sqrt[d]{a})$. Hint: Consider $N_{K/E}(\sqrt[n]{a})\in E$.
Attempt:
I proved that the norm is in $E$ and if $E\supset\mathbb{Q}(\sqrt[d]{a})$ I could prove that $E=\mathbb{Q}(\sqrt[d]{a})$, but I can't prove that $E\supset\mathbb{Q}(\sqrt[d]{a})$.
Let us use multiplicativity of the norm. Write $z=N_{K/E}(\root n\of a)\in E$. Then $$z^n=N_{K/E}((\root n\of a)^n)=N_{K/E}(a)=a^{n/d},$$ because $N_{K/E}(y)=y^{n/d}$ for all $y\in E$. But $E$ is a subfield of the reals, and in the reals the only solutions of $z^n=a^{n/d}$ are $(\pm) \root d\of a,$ where the minus sign is a live possibility only when $n$ is even.
So $z=\pm \root d\of a$. This implies that $\Bbb{Q}(\root d\of a)\subseteq E$. It sounds like you know how to complete the argument.