I tried to figure out an other expression of the product $$\prod_{i=1}^n (x-a_i)$$ and I ended up to the expression $$\sum_{i=0}^n c_i x^i$$ where $$ c_i = \left(-1\right)^{\delta_{n-i\pmod 2} (\{1\})} \sum_{j_k=0, \, j_k \neq k_l,\forall l\neq k \,|l,k\in\{i\}_{i=0}^n}^n\left(\prod_{k=0}^{n-i} a_{j_k}\right)$$ where $$\{ a_{j_k}\}_{j_k=1}^n = \{ a_i\}_{i=1}^n and \, a_{j_0}=1 $$ $$ for \, \, some \, \, integers \, \, \{a_i\}_{i=1}^n \, \, and \, \, a \, \, natural \,\, number \,\, n$$ Is it true?
(Excuse me but I haven't learned to use LaTeX so the formula might be difficult to read. Thank you for your time in advance! )
It is difficult to write that product as a polynomial in the $\{1,\dots , x^n, \dots\}$ base. In fact the complexity of the problem is the same to compute the inverse of a specific matrix.
If you write that product as $\sum_{i=0}^nc_i x^i$ for some appropriate $c_i$, then that coefficients must satisfy the linear system
$\begin{bmatrix} 1 & 0 & \dots & 0\\ 1 & a_1 & \dots & a_1^n \\ \dots \\ 1 & a_n & \dots & a_n^n\\ \end{bmatrix}\begin{bmatrix}c_0\\ \vdots \\ c_n \end{bmatrix}= \begin{bmatrix}\prod_{i=1}^na_i\\ 0 \\ \vdots \\ 0 \end{bmatrix}$
And so the solution is
$\begin{bmatrix}c_0\\ \vdots \\ c_n \end{bmatrix}= \begin{bmatrix} 1 & 0 & \dots & 0\\ 1 & a_1 & \dots & a_1^n \\ \dots \\ 1 & a_n & \dots & a_n^n\\ \end{bmatrix}^{-1} \begin{bmatrix}\prod_{i=1}^na_i\\ 0 \\ \vdots \\ 0 \end{bmatrix}$