I am reading Proposition 10.3 of Neukirch which I have appended below:
Proposition 10.3 (Neukirch): Let $n = \prod_{p} p^{\nu_p}$ be the factorisation of the positive integer $n$ into prime numbers, and for every prime number $p$ denote by $f_p$ the multiplicative order of $p$ mod $n/p^{\nu_p}$. Then one has in $\Bbb{Q}(\zeta_n)$ the factorisation $p = (\mathfrak{p}_1\ldots \mathfrak{p}_r) ^{\varphi(p^{\nu_p})} \pmod{p}$ where the $\mathfrak{p}_i$ are distinct prime ideals of $\mathcal{o} = \Bbb{Z}[\zeta_n]$ lying over $(p)$, all of ramification index $f_p$.
The idea of the proof is that it will suffice to show that $\phi_n(X)$, the minimal polynomial of $\zeta_n$ factors as
$\phi_n(X) = (p_1(X)\ldots p_r(X))^{\varphi(p^{\nu_p})} \pmod{p}.$
Now let us write $n = p^{\nu_p}m$ with $(m,p) = 1$. I get how he deduces the congruence
$\phi_n(X) = \phi_m(X)^{\varphi(p^{\nu_p})} \pmod{p}.$
The part where I don't understand next is where he says:
Observing that $f_p$ is the smallest positive integer such that $p^{f_p} \equiv 1 \pmod{m}$, it is obvious that this congruence reduces us to the case where $p$ does not divide $n$, and hence $\varphi(p^{\nu_p}) = \varphi(1) = 1.$
Why is it that in the case $p|n$ then the result follows? It seems to me he is claiming that when $p|n$ we have the polynomial $x^m-1$ having no multiples roots mod $p$. I don't understand the link between $p$ multiplicative order $f_p$ mod $m$ and the separability of $x^m - 1$ in the case $p|n$
Thanks.