Let $n$ be a natural number and let $E_n=e^{2\pi i/n}$. Then the polynomial $x^n - 1$ splits in $\mathbb{Q}(E_n)$ as
$x^n - 1 = (x - 1)(x - E_n)\cdots(x-E_n^{n-1})$, which yields:
$1+x+\cdots+x^{n-1} = \frac{x^n-1}{x-1} = (x - E_n)(x - E_n^2)\cdots(x-E_n^{n-1})$
and so we get the following result.
$\prod_{j=1}^{n-1}(1-E_n^j) = n$
Now let $n = p^m$, for some natural number $m$. Thus,
$p^m = \prod_{j=1}^{p^m-1}(1-E_{p^m}^j) = \prod_{(j, p^m)=1}(1-E_{p^m}^j)\prod_{(j, p^{m-1})=1}(1-E_{p^{m-1}}^j)\cdots\prod_{(j, p)=1}(1-E_p^j)$
Now look at the right side of the last equation. Each product $\prod_{(j, p^r)=1}(1-E_{p}^j)$ is greater than 1, and there are a total of $m$ factors of this form. Since the product of these $m$ factors is $p^m$, then each factor must be equal to $p$.
Therefore, $\prod_{(j, p^r)=1}(1-E_{p^r}^j) = p$.