Find the product.

$\begin{aligned} P &= \lim_{n \to \infty} \prod_{k=1}^n \frac {(2k)(2k+2)}{(2k+1)^2} \\ &= \lim_{n \to \infty} \frac {2^{2n}n!(n+1)!}{\color{#3D99F6}(2n+1)!!^2} \\ &= \lim_{n \to \infty} 2^{2n}n!(n+1)! \cdot \color{#3D99F6} \left( \frac {2^{n}n!}{(2n+1)!} \right)^2 \\ &= \lim_{n \to \infty} \frac {2^{4n}n!^3(n+1)!}{(2n+1)!^2} & \small \color{#3D99F6} \text{By Stirling's formula: } n! \sim \sqrt{2\pi} n^{n+\frac 12} e^{-n} \\ &= \lim_{n \to \infty} \frac {2^{4n}(\sqrt{2\pi} n^{n+\frac 12} e^{-n})^3\sqrt{2\pi} (n+1)^{n+\frac 32} e^{-n-1}}{(\sqrt{2\pi} (2 n+1)^{2n+\frac 32} e^{-2n-1})^2} \\ &= \lim_{n \to \infty} \frac {2^{4n}(\sqrt{2\pi} n^{n+\frac 12} e^{-n})^3\sqrt{2\pi} n^{n+\frac 32} e^{-n-1}}{\left(\sqrt{2\pi} (2 n)^{2n+\frac 32} {\color{#3D99F6}\left(1+\frac 1{2n}\right) ^{2n+\frac 32}} e^{-2n-1}\right)^2} & \small \color{#3D99F6} \text{Note that } \lim_{n \to \infty} \left(1+\frac 1{2n}\right) ^{2n+\frac 32} = e \\ &= \lim_{n \to \infty} \frac {2^{4n+2}\pi^2 n^{4n+3}e^{-4n-1}}{2^{4n+4}\pi n^{4n+3}e^{-4n-1}} \\ & = \frac \pi 4 \end{aligned}$

$\implies \lfloor 10^{10}P \rfloor = \boxed{7853981633}$