\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{2016^2}\right)\)
\(=\left[\left(1^2\right)-\left(\dfrac{1}{2}\right)^2\right]\left[\left(1^2\right)-\left(\dfrac{1}{3}\right)^2\right]...\left[\left(1\right)^2-\left(\dfrac{1}{2016}\right)^2\right]\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1+\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2015}{2016}\cdot\dfrac{2017}{2016}\)
\(=\dfrac{1}{2}\cdot\dfrac{2017}{2016}=\dfrac{2017}{4032}\)
\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{2016^2}\right)\)
\(=\left[\left(1\right)^2-\left(\dfrac{1}{2}\right)^2\right]\left[\left(1\right)^2-\left(\dfrac{1}{3}\right)^2\right]...\left[\left(1\right)^2-\left(\dfrac{1}{2016}\right)^2\right]\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1+\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2015}{2016}\cdot\dfrac{2017}{2016}\)
\(=\dfrac{1}{2}\cdot\dfrac{2017}{2016}=\dfrac{2017}{4032}\)