I'm reading a solution to a problem and in it they use this fact, It seems intuitive that it should be true, but I dont see how you could show it.
$K$ is a number field of degree $n$, and $\alpha\in K$. $T_\alpha:K\to K$ is a linear transformation from $K$ as a $\mathbb{Q}$ vector space corresponding to multiplication by $\alpha$. Then, if $[\mathbb{Q}(\alpha):\mathbb{Q}]=d$, we get $[K:\mathbb{Q}(\alpha)]=\frac{n}{d}$.
How can we show that $\det(T_\alpha)=\det(T_\alpha|_{\mathbb{Q}(\alpha)})^{n/d}$ and that $N_{K/\mathbb{Q}}(\alpha)=(N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha))^{n/d}$
Take a basis $$K = \sum_{j=1}^{n/d} b_j \Bbb{Q}(\alpha)$$
Each subspace $b_j \Bbb{Q}(\alpha)$ is sent to itself by the linear map $T_\alpha:x\to \alpha x$, thus $$\det(T_\alpha) = \prod_{j=1}^{n/d} \det(T_\alpha|_{b_j \Bbb{Q}(\alpha)})$$ Finally $$\det(T_\alpha|_{b_j \Bbb{Q}(\alpha)})=\det(T_\alpha|_{\Bbb{Q}(\alpha)})$$