Telescopic Product involving Complex Numbers

Question

If $$ \left(1 + \frac{1}{1^2}\right) \left(1 + \frac{1}{(1 + i)^2}\right) \left(1 + \frac{1}{(1 + 2i)^2}\right) \cdots \left(1 + \frac{1}{(1 + (n-1)i)^2}\right) = \frac{10 + 8i}{1 + 8i} $$ find $n$

I came across the above problem a while ago, and the look and feel of it hints at a possible telescopic sum.

I tried to write the term inside the nth bracket in terms of the $(n-1)$-th bracket, but it didn't really help.

Could anyone please point me in the right direction, and tell me how to approach or solve the question?

Thanks a lot.

Answer 1

$$1+\frac1{(1+ki)^2}=\frac{(1+ki)^2-i^2}{(1+ki)^2} =\frac{(1+(k-1)i)(1+(k+1)i)}{(1+ki)^2}$$ etc.