While working with divisors in cyclotomic extensions of $\mathbb{Q},$ I came across this identity:
Given a prime $p$ and an integer $m$ such that $(p,m) =1$, let $f$ be the smallest such the $p^f \equiv 1 \text{ mod } m$ where $p,m,f \in \mathbb{Z}_{\geq 2}.$
How can I prove that $$\prod_{i=1}^{\phi(m)}(1 - \frac{\chi_i(p)}{p^s}) =(1 - \frac{1}{p^{fs}})^{\frac{\phi(m)}{f}}, $$ for any real number $s>1$ where the $\chi_i$ denote the $\phi(m)$ Dirichlet characters mod $m$ ?
I tried grouping the factors on the left into $\frac{\phi(m)}{f}$ groups of $f$ factors so each group product gives me one copy $(1 - \frac{1}{p^{fs}})$ but I have not yet figured out how to group the characters in order to get this.
First use that, $$ X^n - 1 = \prod_{\zeta} \left( X - \zeta\right)$$ Setting $X = p^s$ and dividing should give, $$ 1 - \frac{1}{p^{ns}} = \prod_{\zeta} \left( 1 - \frac{\zeta}{p^s} \right) $$ Then you use the fact that a character modulo $n$ produces the roots of unity (what we are calling $\zeta$, above) as its output values. The rest of the argument is a counting argument were you need to count how many times a Dirichlet character takes on a particular root of unity.