An insane product Production!

Given $p(x)\in\mathbb{Z}[x]$ , the polynomial

$q(x_1, \ldots, x_m) = \prod_{i = 1}^{m}p(x_{i})$ , is a symmetric polynomial in $\mathbb{Z}[x_1,\ldots,x_m]$ . By the fundamental theorem of symmetric polynomials $q(x_1,\ldots,x_m) = r(s_1,\ldots,s_m)$ where the $s_i$ are the elementary symmetric polynomials , and $r \in\mathbb{Z}[x_1,\ldots, x_m]$ (in fact the coefficients of $r$ are uniquely given by the coefficients of $q$ ). Now if $f_1(x) = x^m - c_1x^{m - 1} + \ldots + ( - 1)^mc_m$ . Then $q(a_1,\ldots,a_m) = r(c_1,\ldots,c_m)$ (by Viète's formulas). This means that $\prod_{i = 1}^{m}p(a_{i}) \in \mathbb{Z}$ for any $p(x)\in\mathbb{Z}[x]$ . $(*)$ Finally evaluating $f_1g_1 + f_2g_2 = 1$ in $a_i$ for $i = 1,\ldots,m$ it follows that $f_2(a_i)g_2(a_i) = 1$ and then

$\prod_{i = 1}^{m}f_2(a_{i})\prod_{i = 1}^{m}g_2(a_{i}) = 1$

by $(*)$ both $\prod_{i = 1}^{m}f_2(a_{i})$ and $\prod_{i = 1}^{m}g_2(a_{i})$ are integers. So $\prod_{i = 1}^{m}f_2(a_{i})$ is a unity of $\mathbb{Z}$ , then it is $1$ or $- 1$ , equivalently

$|\prod_{i = 1}^{m}f_2(a_{i})| = 1$ but this is just $|\prod_{i = 1}^{m}\prod_{j = 1}^{n}(a_{i} - b_{j})| = 1$