jmc

Let $y=\frac{1}{1-x}.$ Solving for $x$ in terms of $y,$ we find $$x=\frac{y-1}{y}.$$Then $${\left(\frac{y-1}{y}\right)}^{2016}+{\left(\frac{y-1}{y}\right)}^{2015}+\cdots +\left(\frac{y-1}{y}\right)+1=0.$$Hence, $$(y-1{)}^{2016}+y(y-1{)}^{2015}+y^2(y-1{)}^{2014}+\cdots +y^{2015}(y-1)+y^{2016}=0.$$This expands as $$\begin{array}{rl}& \left(y^{2016}-2016y^{2015}+\left(\genfrac{}{}{0ex}{}{2016}{2}\right)y^{2014}-\cdots \right)\\ & +y\left(y^{2015}-2015y^{2014}+\left(\genfrac{}{}{0ex}{}{2015}{2}\right)y^{2013}-\cdots \right)\\ & +y^2\left(y^{2014}-2014y^{2013}+\left(\genfrac{}{}{0ex}{}{2014}{2}\right)y^{2012}-\cdots \right)\\ & +\cdots \\ & +y^{2015}(y-1)+y^{2016}=0.\end{array}$$The coefficient of $y^{2016}$ is 2017. The coefficient of $y^{2015}$ is $$-2016-2015-\cdots -2-1=-\frac{2016\cdot 2017}{2}=-2033136.$$The coefficient of $y^{2014}$ is $$\left(\genfrac{}{}{0ex}{}{2016}{2}\right)+\left(\genfrac{}{}{0ex}{}{2015}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{2}{2}\right).$$By the Hockey Stick Identity, $$\left(\genfrac{}{}{0ex}{}{2016}{2}\right)+\left(\genfrac{}{}{0ex}{}{2015}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{2}{2}\right)=\left(\genfrac{}{}{0ex}{}{2017}{3}\right)=1365589680.$$The roots of the polynomial in $y$ above are $y_k=\frac{1}{1-x_k}$ for $1\le k\le 2016,$ so by Vieta's formulas, $$y_1+y_2+\cdots +y_{2016}=\frac{2033136}{2017}=1008,$$and $$y_1{y}_2+y_1{y}_3+\cdots +y_{2015}{y}_{2016}=\frac{1365589680}{2017}=677040.$$Therefore, $$\begin{array}{rl}& \frac{1}{(1-x_1{)}^2}+\frac{1}{(1-x_2{)}^2}+\cdots +\frac{1}{(1-x_{2016}{)}^2}\\ & =y_1^2+y_2^2+\cdots +y_{2016}^2\\ & =(y_1+y_2+\cdots +y_{2016}{)}^2-2(y_1{y}_2+y_1{y}_3+\cdots +y_{2015}{y}_{2016})\\ & =1008^2-2\cdot 677040\\ & =\boxed{-338016}.\end{array}$$