TITLE: I want to show that the polynomial below has integer coefficients.
QUESTION [12 upvotes]: Edit: According to the suggestions in comments, this only holds when f(x) is monic.
Let $p$ be a prime number and $f(x) = \sum_{i=0}^{p} a_{i}x^{i}$ be a monic polynomial with $a_{0}, a_{1}, \ldots, a_{p} \in\mathbb{Z}$. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{p}$ be its roots.
I am trying to show that the polynomial $$g(x) = \frac{x^{p} - \alpha_{1}^{p}}{x - \alpha_{1}} \cdot \frac{x^{p} - \alpha_{2}^{p}}{x - \alpha_{2}} \cdots \frac{x^{p} - \alpha_{p}^{p}}{x - \alpha_{p}}$$ has integer coefficients.
My attempt: I managed to show that the polynomial $$h(x) = (x^{p} - \alpha_{1}^{p})\cdot (x^{p} - \alpha_{2}^{p})\cdots (x^{p} - \alpha_{p}^{p})$$ is a polynomial with integer coefficients. This is because coefficient of $f(x)$ are values of elementary symmetric polynomial in $p$ variables, evaluated at the roots $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{p}$ and coefficient of $h(x)$ are values of same elementary symmetric polynomial evaluated at $\alpha_{1}^{p}, \ldots, \alpha_{p}^{p}$. Then I used Fundamental Theorem of Symmetric Polynomials.
My strategy was to show that when I divide $h(x)$ by $(x-\alpha_{1})\cdots (x-\alpha_{p})$, I get integer polynomial as a result. I thought of doing this partially because it might help me find the coefficents of $g(x)$ too. But now, I am struggling to show that.
Are there any other strategy to this problem (preferably that can help me isolate the coefficients of $g(x)$)?

REPLY [6 votes]: If $\,a_p \ne 1$ the proposition does not hold true in general, as noted in a comment. Otherwise if $\,a_p=1\,$, let $\,\displaystyle P_n(z) = \prod_{k=1}^p(z- \alpha_k^n)\,$, and $\,\displaystyle q(x)=\frac{P_n(x^n)}{P_1(x)}\,$.
$T_n(x)=P_n(x^n)$ is symmetric in $\,\alpha_k\,$, so its coefficients are symmetric functions of the coefficients $\,a_i\,$ and therefore integers themselves. Moreover, $\,T_n(\alpha_i)=P_n(\alpha_i^n)=0\,$, so $\,P_1(x) \mid T_n(x)\,$, and $\,\displaystyle q(x)=\frac{T_n(x)}{P_1(x)}\,$ has integer coefficients as the quotient of two monic polynomials with integer coefficients.
$P_n(z)\,$ can be explicitly determined either by expanding the product and expressing the coefficients in terms of the elementary symmetric polynomials in $\,\alpha_k\,$, or by using polynomial resultants $\,P_n(z) = \text{res}\left(P_1(x), z-x^n, x\right)\,$.
The proposition in OP's question follows for $\,n=p\,$, and it holds true whether $\,p\,$ is a prime or not.