Let $x_1,$ $x_2,$ $\dots,$ $x_{2016}$ be the roots of
\[x^{2016} + x^{2015} + \dots + x + 1 = 0.\]Find
\[\frac{1}{(1 - x_1)^2} + \frac{1}{(1 - x_2)^2} + \dots + \frac{1}{(1 - x_{2016})^2}.\]
Let $y = \frac{1}{1 - x}.$  Solving for $x$ in terms of $y,$ we find
\[x = \frac{y - 1}{y}.\]Then
\[\left( \frac{y - 1}{y} \right)^{2016} + \left( \frac{y - 1}{y} \right)^{2015} + \dots + \left( \frac{y - 1}{y} \right) + 1 = 0.\]Hence,
\[(y - 1)^{2016} + y (y - 1)^{2015} + y^2 (y - 1)^{2014} + \dots + y^{2015} (y - 1) + y^{2016} = 0.\]This expands as
\begin{align*}
&\left( y^{2016} - 2016y^{2015} + \binom{2016}{2} y^{2014} - \dotsb \right) \\
&+ y \left( y^{2015} - 2015y^{2014} + \binom{2015}{2} y^{2013} - \dotsb \right) \\
&+ y^2 \left( y^{2014} - 2014y^{2013} + \binom{2014}{2} y^{2012} - \dotsb \right) \\
&+ \dotsb \\
&+ y^{2015} (y - 1) + y^{2016} = 0.
\end{align*}The coefficient of $y^{2016}$ is 2017.  The coefficient of $y^{2015}$ is
\[-2016 - 2015 - \dots - 2 - 1 = -\frac{2016 \cdot 2017}{2} = -2033136.\]The coefficient of $y^{2014}$ is
\[\binom{2016}{2} + \binom{2015}{2} + \dots + \binom{2}{2}.\]By the Hockey Stick Identity,
\[\binom{2016}{2} + \binom{2015}{2} + \dots + \binom{2}{2} = \binom{2017}{3} = 1365589680.\]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,
\[y_1 + y_2 + \dots + y_{2016} = \frac{2033136}{2017} = 1008,\]and
\[y_1 y_2 + y_1 y_3 + \dots + y_{2015} y_{2016} = \frac{1365589680}{2017} = 677040.\]Therefore,
\begin{align*}
&\frac{1}{(1 - x_1)^2} + \frac{1}{(1 - x_2)^2} + \dots + \frac{1}{(1 - x_{2016})^2} \\
&= y_1^2 + y_2^2 + \dots + y_{2016}^2 \\
&= (y_1 + y_2 + \dots + y_{2016})^2 - 2(y_1 y_2 + y_1 y_3 + \dots + y_{2015} y_{2016}) \\
&= 1008^2 - 2 \cdot 677040 \\
&= \boxed{-338016}.
\end{align*}