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)^{-1}\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.

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.

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 $u,v$. Note that similarity respects the sum and product, 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)^{-1}\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.

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)^{-1}\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.