Why is it that for real numbers $0<x_i<1$ $(1-x_1)(1-x_2)\cdots(1-x_n)(1-x_{n+1})=1-x_1-x_2-\cdots-x_n-x_{n+1}+x_{n+1}(x_1+\cdots+x_n)$?
The most I was able to do was factor out the $x$ terms individually, \begin{align*} &= (1-x_1)(1-x_2)\cdots (1-x_n)(1-x_{n+1}) \\ &= x_1(\frac{1}{x_1}-1)x_2(\frac{1}{x_2}-1)\cdots x_n(\frac{1}{x_n}-1)x_{n+1}(\frac{1}{x_{n+1}}-1) \\ &= (x_1\cdots x_{n+1})(\frac{1}{x_1}-1)\cdots (\frac{1}{x_{n+1}}-1)\end{align*}
The identity you give is true for $n=1$:
$$(1-x_1)(1-x_2)=1-x_1-x_2+x_2x_1$$
but for $n>1$, it is erroneous in general.
For example, in the case $n=2$, if $x_1=x_2=x_3=1/2$:
$$\underbrace{(1-x_1)(1-x_2)(1-x_3)}_{\tfrac{1}{8}}\ne \underbrace{1-x_1-x_2-x_3+x_3(x_1+x_2)}_0$$