Laurent polynomials: what is the correspondence here? Given a Laurent polynomial $f$, denote the number of terms by $\#f$ and let $\widehat{CT}(f)$ stand for the value of the constant term in $f$. For example, if $f(x,y)=2-\frac{y}x-\frac{x}y$ then $\#f=3$ and $\widehat{f}=2$.
Define the two functions
\begin{align*}
P_n=\prod_{i=1}^n\prod_{j=1}^{i-1}\left(1-\frac{x_i}{x_j}\right)\left(1-\frac{x_j}{x_i}\right) \qquad \text{and} \qquad
Q_n=\prod_{j=1}^{n-1}\prod_{j=i}^{n-1}\left(1-\prod_{k=i}^jy_k\right). \end{align*}

QUESTION 1. I know $\widehat{CT}(P_n)=\#Q_n$. Can you establish a direct combinatorial connection between the two objects?


QUESTION 2. The same question for $\#P_n=\#(Q_n^2)$.

Example 1. $P_2=2-\frac{x_1}{x_2}-\frac{x_2}{x_1}$ and $Q_2=1-y_1 \implies \widehat{CT}(P_2)=\#Q_2=2$ and $\#P_2=\#(Q_2^2)=3$.
Example 2. Keep in mind that $\widehat{CT}(P_n)=\#Q_n=n!$. The first few values of
$\#P_n$: $1, 3, 19, 201, 2961, 56183, 1392385, \dots$
 A: Notice that $f$ is a homogeneous Laurent polynomial of degree zero. We can set $x_n=1$ and that will not affect $\widehat{CT}(f)$ or $\#f$. Suppose that we have the relations $x_i=\prod_{j=1}^{i}y_j$ for each $1\le i \le n-1$. This change of variables induces a bijection between Laurent monomials that sends 1 to 1. In particular computing $\widehat{CT}(f)$ or $\# f$ isn't affected by the change of variables.
It is easy to check that
$$P(x_1,x_2,\dots,x_{n-1})=Q(y_1,y_2,\dots,y_{n-1})Q(y_1^{-1}, y_2^{-1}, \dots,y_{n-1}^{-1}). \tag{*}$$
Now notice that $Q(y_1,\dots,y_{n})$ is equal to the product of the Vandermonde $\prod (x_i-x_j)$ by a certain monomial, and so in particular it has all coefficients $\pm1$. Therefore, since the constant term of the right hand side of $(*)$ is equal to the sum of the squares of the coefficients of $Q$, it is equal to $\#Q$.
For question 2, notice that $Q(y_1^{-1}, \dots, y_{n-1}^{-1})$ is the same as $Q(y_1,\dots,y_{n-1})$ up to multiplication by a Laurent monomial, therefore the number of terms in the right hand side of $(*)$ is the same as $\# Q^2$, from which your observation follows.
