I was reading Marcus' Number Fields and came across a line saying "we know $(1 - \omega)^{\phi(m)} = p\mathbb{Z}[\omega]$" where $\omega = e^{2\pi i/m}$ and $m = p^r$. Earlier on in the book it proves that $\prod_{k} (1 - \omega^k) = p$ where the product runs over all k, $1 \leq k \leq m$ such that $p \nmid k$.
But these statements aren't equivalent and I can't see how the proof of the second statement could be reworked to fit the first.
Any help would be greatly appreciated.
It follows from the fact that $\frac{1-\omega}{1-\omega^k}$ is a unit in $\Bbb{Z}[\omega]$ whenever $k$ is coprime to $p$. Of course we have $$\frac{1-\omega}{1-\omega^k}\cdot\frac{1-\omega^k}{1-\omega}=1,$$ in $\Bbb{Q}(\omega)$, where $$\frac{1-\omega^k}{1-\omega}=1+\omega+\cdots+\omega^{k-1}\in\Bbb{Z}[\omega],$$ so it remains to show that $\frac{1-\omega}{1-\omega^k}\in\Bbb{Z}[\omega]$. For the automorphism $\varphi$ of $\Bbb{Z}[\omega]$ mapping $\omega$ to $\omega^k$ you have $$\frac{1-\omega}{1-\omega^k}=\varphi\left(\frac{1-\omega^l}{1-\omega}\right),$$ for some $l$ coprime to $p$, where $$\frac{1-\omega^l}{1-\omega}=1+\omega+\ldots+\omega^{l-1}\in\Bbb{Z}[\omega],$$ and hence also $\frac{1-\omega}{1-\omega^k}\in\Bbb{Z}[\omega]$.