If $\left(1 + i\right) \left(1 + 2i\right) \left(1 + 3i\right) \cdots \left(1 + ni\right) = x + iy$ $\;$ show that $\;$ $2 \cdot 5 \cdot 10 \cdots \left(1 + n^2\right) = x^2 + y^2$
Given: $\;$ $\left(1 + i\right) \left(1 + 2i\right) \left(1 + 3i\right) \cdots \left(1 + ni\right) = x + iy$
i.e. $\;$ $\left|\left(1 + i\right) \left(1 + 2i\right) \left(1 + 3i\right) \cdots \left(1 + ni\right)\right| = \left|x + iy\right|$
i.e. $\;$ $\left|1 + i\right| \; \left|1 + 2i\right| \; \left|1 + 3i\right| \cdots \left|1 + ni\right| = \left|x + iy\right|$
i.e. $\;$ $\sqrt{1^2 + 1^2} \; \sqrt{1^2 + 2^2} \; \sqrt{1^2 + 3^2} \cdots \sqrt{1^2 + n^2} = \sqrt{x^2 + y^2}$
i.e. $\;$ $\sqrt{2} \; \sqrt{5} \; \sqrt{10} \cdots \sqrt{1 + n^2} = \sqrt{x^2 + y^2}$ $\;\;\; \cdots \; (1)$
Squaring both sides of equation $(1)$ we have,
$2 \cdot 5 \cdot 10 \cdots \left(1 + n^2\right) = x^2 + y^2$
Hence proved.