How to prove that $\Pi_{1 \leq j \leq p^n, p \nmid j}((\zeta_{p^n}^j)^{p^{n-1}}-1) = p^{p^{n-1}}$?

Question

How do I prove that $\Pi_{1 \leq j \leq p^n, p \nmid j}((\zeta_{p^n}^j)^{p^{n-1}}-1) = p^{p^{n-1}}$? I calculated some examples in Wolfram Alpha where the results were of this specific form, and it also has to be of this specific form in my calculation in the context of finding the discriminant of $d(1,\zeta_{p^n},...,\zeta_{p^n}^{p^n-1})$ (I know that the discriminant has to look like $(-1)^k * p^{p^{n-1}*(np-n-1)}, k \in N$ and I already got the term of $(-1)^k * p^{p^{n-1}*(np-n)}$ with the numerator of a fraction, where $\Pi_{1 \leq j \leq p^n, p \nmid j}((\zeta_{p^n}^j)^{p^{n-1}}-1)$ is left in the denominator of a fraction), so how do I show this identity?

Answer 1

I’m assuming $p^n >2$.

When $1\leq j \leq p^n$ is coprime to $p$, $(\zeta_{p^n}^j)^{p^{n-1}}=\zeta_p^{j\mathrm{ mod }p}$, and since $\{1 \leq j \leq p^n,\, j \wedge p=1\}$ contains $p^{n-1}$ elements with a fixed nonzero residue mod $p$, your product is the $p^{n-1}$-th power of $\prod_{1 \leq j <p}{(\zeta_p^j-1)}$, which is the evaluation at $x=1$ of $\frac{(-1)^{p-1}}{x-1}\prod_{z^p=1}{(x-z)}=\frac{(-1)^{p-1}(x^p-1)}{x-1}$, and at $x=1$ this is $p(-1)^{p-1}$.

So your overall product is $((-1)^{p-1}p)^{p^{n-1}}=(-1)^{p^{n-1}(p-1)}p^{p^{n-1}}=p^{p^{n-1}}$.