$\DeclareMathOperator\CT{CT}$
Let $\CT\_t(f(t))$ denote the constant term of the Laurent polynomial of $f(t)$.


Define the two functions $F(x\_1,\dots,x\_n)$ and $G(y)$ by
$$F:=\prod\_{i=1}^nx\_i^{-1}(1-x\_i)^{-2}\prod\_{1\leq i<j\leq n}
(1-x\_i-x\_j)^{-1} \qquad \text{and} \qquad 
G:=n!\cdot y^{-n}e^{(n+1)y+y^2/2}.$$
I like to ask:



> 
> **QUESTION.** Is the following true? It would be great if there is a direct way to compare these two.
> $$\CT\_{x\_1}\CT\_{x\_2}\cdots\CT\_{x\_n}\left(F(x\_1,\dots,x\_n)\right)=\CT\_y\left(G(y)\right).$$
> 
> 
> 


**NOTE 1.** The sequence on the right-hand side is available at the [OEIS as A301741](https://oeis.org/A301741) with an explicit evaluation.


**NOTE 2.** Incidentally, we also have (a consequence of Han's formula [proved here](https://arxiv.org/pdf/0808.0928.pdf))
$$\CT\_{x\_1}\CT\_{x\_2}\cdots\CT\_{x\_n}\left(F(x\_1,\dots,x\_n)\right)=
\sum\_{\lambda\vdash n}f^{\lambda}\prod\_{u\in\lambda}
\frac{(n+2)^{h\_u}+n^{h\_u}}{(n+2)^{h\_u}-n^{h\_u}};$$
where $h\_u$ is the *hook-length* of cell $u$ (in the Young diagram of $\lambda$) and $f^{\lambda}$ is the *number of Standard Young Tableau of shape $\lambda$* ([given by the hook-length formula](https://en.wikipedia.org/wiki/Hook_length_formula)).


**NOTE 3.** A cute analogue: let $f:=\prod\_{i=1}^nx\_i^{-1}(1-x\_i)^{-1}\prod\_{1\leq i<j\leq n}(1+x\_i+x\_j)$ and $g:=n!\cdot y^{-n}e^{ny-y^2/2}$. Then,
$$\CT\_{x\_1}\CT\_{x\_2}\cdots\CT\_{x\_n}\left(f(x\_1,\dots,x\_n)\right)
=\CT\_y(g(y)).$$
**Proof.** Fedor's reasoning applies (it'd be nice to employ Richard's too)
\begin{align\*}
{\rm CT}\, f&=
[x\_1\ldots x\_n] \prod\_i (1-x\_i)^{-1}\prod\_{i<j}(1+x\_i+x\_j) \\
&=[x\_1\ldots x\_n]\prod\_i\exp(x\_i)\prod\_{i<j}\exp(x\_i+x\_j-x\_ix\_j)\\
&=[x\_1\ldots x\_n] \exp\left(\sum x\_i+\sum\_{i<j}(x\_i+x\_j-x\_ix\_j)\right)\\
&=[x\_1\ldots x\_n]\exp\left(n\cdot S-S^2/2\right).
\end{align\*}



Extracting constant terms: is there a direct way?
For power series $u(x\_1,\ldots,x\_n),v(x\_1,\ldots,x\_n)$ call $u,v$ similar and write $u\sim v$ if all monomials $\prod x\_i^{c\_i}$ with $c\_i\in \{0,1\}$ have equal coefficients in $u,v$. In other words, if $u$ is congruent to $v$ modulo the ideal generated by $x\_i^2$'s. Note that this similarity respects addition and multiplucation, and that $(1-x\_i)^{-1}\sim \exp(x\_i)$ and $(1-x\_i-x\_j)^{-1}\sim 1+x\_i+x\_j+2x\_ix\_j\sim\exp(x\_i+x\_j+x\_ix\_j)$. Thus
\begin{align\*}
{\rm CT}\, F&=
[x\_1\ldots x\_n] \prod\_i (1-x\_i)^{-2}\prod\_{i<j}(1-x\_i-x\_j)^{-1}\\&=
[x\_1\ldots x\_n]\prod\_i\exp(2x\_i)\prod\_{i<j}\exp(x\_i+x\_j+x\_ix\_j)\\&=[x\_1\ldots x\_n] \exp\left(
\sum 2x\_i+\sum\_{i<j}(x\_i+x\_j+x\_ix\_j)
\right)\\
&=[x\_1\ldots x\_n]\exp\left((n+1)S+S^2/2\right),
\end{align\*}
where $S=x\_1+\ldots+x\_n$ (since $S^2/2\sim \sum\_{i<j} x\_ix\_j$). Now if we expand $\exp\left((n+1)S+S^2/2\right)$ as a power series in $S$, we get $[x\_1\ldots x\_n]S^n=n!$ and $[x\_1\ldots x\_n]S^k=0$ for $k\ne n$, thus your identity.


