In the Problems from the Book by Titu Andreescu, there is a proof of Example 9 on page 494 with the following:
Example 9. Let $f$ be a monic polynomial with integer coefficients and let $p$ be a prime number. If $f$ is irreducible in $\mathbb{Z}[x]$ and $\sqrt[p]{(-1)^{\deg f}f(0)}$ is irrational, then $f(X^p)$ is also irreducible in $\mathbb{Z}[x]$.
Solution. Consider $\alpha$ a complex zero of $f$ and let $n=\deg f$ and $g(X)=X^p$ and $h=g-\alpha$. Using previous results, it suffices to prove that $h$ is irreducible in $\mathbb{Q}[\alpha][X]$. Because $\mathbb{Q}[\alpha]$ is a subfield of $\mathbb{C}$, it suffices to prove that $\alpha$ is not the $p-th$ power of an element of $\mathbb{Q}[\alpha]$. Suppose there is $u\in \mathbb{Q}[x]$ of degree at most $n-1$ such that $\alpha=u^p(\alpha)$ . Let $\alpha_1, \alpha_2, \dots, \alpha_n$ be the zeroes of $f$. Because $f$ is irreducible and $\alpha$ is one of its zeroes, $f$ is the minimal polynomial of $\alpha$, so $f$ must divide $u^p(X)-X$. Therefore $\alpha_1\cdot\alpha_2\cdots\alpha_n=(u(\alpha_1)\cdot u(\alpha_2)\cdots u(\alpha_n))^p.$ Finally, using the fundamental theorem of symmetric polynomials, $u(\alpha_1)\cdot u(\alpha_2)\cdots u(\alpha_n)$ is rational. But $\alpha_1\cdot \alpha_2\cdots \alpha_n=(-1)^nf(0)$, implies $\sqrt[p]{(-1)^nf(0)} \in \mathbb{Q}$, a contradiction.
My question is how does the fundamental theorem of symmetric polynomials imply that $u(\alpha_1)\cdot u(\alpha_2)\cdots u(\alpha_n)$ is rational?
The theorem says that any symmetrical polynomial can be uniquely expressed as a polynomial in elementary symmetric polynomials (one formulation is here The Fundamental Theorem of Symmetric Polynomials), but how does that apply in this case and leads to rationality?