Theorem 30 in Marcus book states that, if $p\in\mathbb Z$ is an odd prime and $q$ is a prime $\neq p$, then, fixing $d$ as a divisor of $p-1$ we have that $q$ is a $d$-th power $\operatorname{mod}q$ iff $q$ splits completely in $F_d$.
Now, we are in $\mathbb Q[\omega]$, where $\omega=e^{\frac{2\pi i}{p}}$, which is a normal extension of $\mathbb Q$ of degree $\varphi(p)=p-1$. Hence $G:=\operatorname{Gal}(\mathbb Q[\omega]|\mathbb Q)$ is a cylic group of order $p-1$. Hence for all $d|p-1$, there exists a unique $H\le G$ of order $(p-1)/d$ and (from Galois theory) it corresponds to a field $K$, with $\mathbb Q\le K\le\mathbb Q[\omega]$ and $[K:\mathbb Q]=d$. We call it $F_d$.
We know that, being $\mathbb Q[\omega]$ normal over $\mathbb Q$, considering the prime $q$ of $\mathbb Q$, it splits in $\mathbb Q[\omega]$ as $q\mathbb Q[\omega]=(Q_1\dots Q_r)^e$ and moreover we have that $ref=\varphi(p)=p-1$, where $e$ and $f$ are respectively the ramification index and the inertial degree of $q$ in $\mathbb Q[\omega]$.
Now the proof begins saying that $q$ splits in $r$ primes in $\mathbb Q[\omega]$ and $f=(p-1)/r$ (which is the order of $q\;\operatorname{mod}p$).
My problem is: why is $f=(p-1)/r$?! I know that if I show $e=1$ I'd conclude because $f=(p-1)/r$ is equivalent to $e=1$ , but I don't know how to show it! Can somebody help me? Thanks a lot!