Let $p$ be an odd prime. $p$ and $(1-\zeta_p)^{p-1}$ are associates in $\mathbb{Z}[\zeta_p]$.

Let $p$ be an odd prime and $\zeta_p$ a primitive $p$th root of unity, that is a $p$th root of unity other than 1. I need to show that $p$ and $(1-\zeta_p)^{p-1}$ are associates in $\mathbb{Z}[\zeta_p]$.

I don't know what I can do about this. Should I try factorizing $p$ into polynomials in $\zeta_p$?

Recall that: $$ \text{Nm}(1-\zeta)=(1-\zeta)\cdots(1-\zeta^{p-1})=\Phi(1)=p, $$ where $\Phi$ is the $p$-th cyclotomic polynomial. So the ideal generated by $p$ is: $$ (p)=(1-\zeta)\cdots(1-\zeta^{p-1}). $$

Lemma. Let $r,s\in\mathbb{Z}$ such that $\text{gcd}(p,rs)=1$. Then: $$ \frac{\zeta^r-1}{\zeta^s-1}\in(\mathbb{Z}[\zeta])^\times $$

Now $1-\zeta$ and $1-\zeta^i$ are easily shown to be associate if $\text{gcd}(p,i)=1$ by the above lemma, so they generate the same ideal, and: $$ (1-\zeta)^{p-1}=(p). $$ So $(1-\zeta)^{p-1}$ and $p$ are associate.

Let $f(x)=(x^p-1)/(x-1)$. By considering $f(1)$, show that $\prod(1-\zeta_p^k)=p$, the product being over $1\le k\le p-1$. Then show that the numbers $1-\zeta_p^k$ are all associates, by showing that both $(1-\zeta_p^k)/(1-\zeta_p)$ and its inverse are algebraic integers.