I am having some troubles in understanding the proof of Theorem 10, Chapter 2 of D. Marcus' book Number fields, which shows that the ring of algebraic integers in the $p^r$-th cyclotomic field $\mathbb{Q}[\omega]$ (with $p$ prime and $\omega=e^{2\pi i/p^r}$) is $\mathbb{Z}[\omega]$.
In the proof, the author uses the fact that $\mathbb{Z}[\omega]=\mathbb{Z}[1-\omega]$ and that $d:=\text{disc}(\omega)=\text{disc}(1-\omega)$, so every element $\alpha$ in the ring $R$ of algebraic integers of $\mathbb{Q}[\omega]$ can be written as $$\alpha=\frac{m_1+m_2(1-\omega)+\dots +m_n(1-\omega)^{n-1}}{d} $$ for suitable integers $m_1,...,m_n\in \mathbb{Z}$ and with $n:=\varphi(p^r)$. We know that $d$ divides $p^{rn}$, so $d$ is a power of $p$.
If $R\ne \mathbb{Z}[\omega]=\mathbb{Z}[1-\omega] $, then there is some $m_j$ which is not divisible by $d$ in $\mathbb{Z}$.
So far I understand the proof. But now the author says that it follows that $R$ contains an element $\beta$ of the form $$\beta=\frac{m_i(1-\omega)^{i-1}+m_{i+1}(1-\omega)^{i}+\dots +m_n(1-\omega)^{n-1}}{p} $$ for a suitable integer $i\le n$, and for some $m_i,...,m_n\in \mathbb{Z}$ with $m_i$ not divisible by $p$.
I don't see how we can construct such an element from the facts given above. Can someone give me a hint on how to argue? Thanks a lot!