When 1 over a square is subtracted from one

$\begin{aligned} &\left( 1 - \dfrac{1}{4} \right) \left( 1 - \dfrac{1}{9} \right) \left( 1 - \dfrac{1}{16} \right) ... \left( 1 - \dfrac{1}{4072324} \right) \\ \\ = \ &\left( 1 - \dfrac{1}{2} \right) \left( 1 + \dfrac{1}{2} \right) \left( 1 - \dfrac{1}{3} \right) \left( 1 + \dfrac{1}{3} \right) \left( 1 - \dfrac{1}{4} \right) \left( 1 + \dfrac{1}{4} \right) ... \left( 1 - \dfrac{1}{2018} \right) \left( 1 + \dfrac{1}{2018} \right) \\ \\ = \ &\left( 1 - \dfrac{1}{2} \right) \left( 1 - \dfrac{1}{3} \right) \left( 1 - \dfrac{1}{4} \right) .... \left( 1 - \dfrac{1}{2018} \right) \ \times \ \left( 1 + \dfrac{1}{2} \right) \left( 1 + \dfrac{1}{3} \right) \left( 1 + \dfrac{1}{4} \right) ... \left( 1 + \dfrac{1}{2018} \right) \\ \\ = \ &\left( \dfrac{1}{2} \right) \left( \dfrac{2}{3} \right) \left( \dfrac{3}{4} \right) ... \left( \dfrac{2017}{2018} \right) \ \times \ \left( \dfrac{3}{2} \right) \left( \dfrac{4}{3} \right) \left( \dfrac{5}{4} \right) ... \left( \dfrac{2019}{2018} \right) \\ \\ = \ &\dfrac{1}{2018} \times \dfrac{2019}{2} \\ \\ = \ &\boxed{\dfrac{2019}{4036}} \end{aligned}$

$\begin{aligned} P & = \left(1 - \frac 14\right) \left(1 - \frac 19\right) \left(1 - \frac 1{16}\right) \cdots \left(1 - \frac 1{4072324}\right) \\ & = \left(1 - \frac 1{2^2}\right) \left(1 - \frac 1{3^2}\right) \left(1 - \frac 1{4^2}\right) \cdots \left(1 - \frac 1{2018^2}\right) \\ & = \prod_{n=2}^{2018} \left(1-\frac 1{n^2}\right) \\ & = \prod_{n=2}^{2018} \frac {n^2-1}{n^2} \\ & = \prod_{n=2}^{2018} \frac {(n-1)(n+1)}{n^2} \\ & = \frac {2017! \cdot 2019!}{2(2018!)^2} \\ & = \boxed{\dfrac {2019}{4036}} \end{aligned}$