The least positive integer $n$ such that $\left( \frac{2i}{1 + i} \right)^n$ is a positive integer is

First simplify the base. $$\frac{2i}{1+i}=\frac{2i(1-i)}{(1+i)(1-i)}$$ $$\frac{2i}{1+i}=i(1-i)=1+i$$ So we need the least positive integer $n$ such that $$(1+i)^n$$ is a positive integer. Write $1+i$ in polar form. $$1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$$ Hence $$ (1+i)^n=(\sqrt{2})^n\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right) $$ For this to be a positive integer, the imaginary part must be $0$ and the value must be positive real. So we need $$\sin\frac{n\pi}{4}=0$$ and $$\cos\frac{n\pi}{4}=1$$ That means $$\frac{n\pi}{4}=2k\pi$$ for some integer $k$. Therefore $$n=8k$$ The least positive value is $8$. Checking the options, $8$ corresponds to option $C$.