I'm trying to convince myself that when $p\nmid m$, $p$ prime, then $\left(\frac{\mathbb{Q}(\zeta_m)/\mathbb{Q}}{p}\right)$ is the map $\zeta_m\mapsto \zeta_m^p$. I'm sure it's both true and obvious, as all the references I've looked at state it without explanation. If $p$ splits completely, so that $N(p)=p$, then $$\left(\frac{\mathbb{Q}(\zeta_m)/\mathbb{Q}}{p}\right)(\zeta_m) \equiv \zeta_m^{N(p)} = \zeta_m^p\bmod{\mathfrak{p}}$$ for each $\mathfrak{p}$ over $p$, so it's clear here. But if $p$ is inert, for example, we get $$\left(\frac{\mathbb{Q}(\zeta_m)/\mathbb{Q}}{p}\right)(\zeta_m)\equiv \zeta_m^{N(p)} = \zeta_m^{p^{f}}\bmod{p},$$ and I just don't see how we get to $\zeta_m^p$.
Edit: I think I see the answer. If $p = \prod \mathfrak{p}_i$, then $\zeta_m^{p^f}\equiv \zeta_m^p\bmod{\mathfrak{p}_i}$ by Fermat's little theorem, so it is $\zeta_m^p\bmod{p}$ as well.
For a general abelian extension $L/K$ of number fields, the Artin element satisfies
$$\left(\frac{L/K}{\mathfrak{p}}\right)(\alpha)\equiv\alpha^{N_{K/\mathbb{Q}}(\mathfrak{p})}\bmod\mathfrak{P},$$
where $\mathfrak{P}$ is a prime lying over $\mathfrak{p}$.
Here, $N(p)=N_{\mathbb{Q}/\mathbb{Q}}(p)=p$, regardless of how $(p)$ splits.