Why is $m \infty$ the conductor of $K = \mathbb{Q}(\zeta_m)/\mathbb{Q}$?

Question

Wouldn't this be saying that for all $p$ dividing $m$, $1 + p^{\operatorname{ord}_p(m)} \mathbb{Z}_p$ is contained in the group of local norms $N_{\mathfrak p/p}(K_{\mathfrak p})$, where $\mathfrak p$ is a prime lying over $p$? Can this fact been seen directly?