Every element of the form $x^n - \beta$ is a norm? Let $F$ be a local $p$-adic field containing the $n$th roots of unity.  The notes I'm reading claim that every element of the form $x^n - \beta$, for $x, \beta \in F$, is a norm from $F(\sqrt[n]{\beta})$.  Why is this?
I know that $$x^n - \beta = \prod\limits_{i=0}^{n-1}(x - \zeta_n^i \sqrt[n]{\beta})$$ so the norm of $x - \sqrt[n]{\beta}$ should be some subproduct of the right hand side.
 A: Since $F$ contains all the $n$th roots of unity, the extension $F(\sqrt[n]{\beta})/F$ is Galois. Any automorphism $\sigma\in\mathrm{Gal}(F(\sqrt[n]{\beta})/F)$ is uniqely determined by its action on $\sqrt[n]{\beta}$. Fix some root $\alpha=\sqrt[n]{\beta}$. Every automorphism $\sigma$ maps it to an element of the form $\zeta^i_n\alpha$, so the map $\sigma\mapsto\frac{\sigma(\alpha)}{\alpha}$ embeds $\mathrm{Gal}(F(\sqrt[n]{\beta})/F)$ into the group of $n$th roots of unity $\mu_n$. Hence, $\mathrm{Gal}(F(\alpha)/F)$ is isomoprhic to some subgroup $\mu_d\subset\mu_n$. Fix some representatives of cosets $\mu_n/\mu_d$: $\xi_1,\dots,\xi_{n/d}$. I claim that the element $\prod\limits_{i=1}^{n/d}(x-\xi_i\alpha)$ has the norm equal to $x^n-\beta$. Indeed, the norm of every multiplier $(x-\xi_i\alpha)$ is equal to $$\prod\limits_{\sigma\in\mathrm{Gal}(F(\alpha)/F)}\sigma(x-\xi_i\alpha)=\prod\limits_{\zeta\in\xi_i \mu_d}(x-\zeta\alpha)$$ so $N_{F(\alpha)/F}\prod\limits_{i=1}^{n/d}(x-\xi_i\alpha)=\prod\limits_{\zeta\in\mu_n}(x-\zeta\alpha)=x^n-\beta$ 
