complex algebra Calculate: \begin{align*}
&\left( 1+\frac{1+i}{2} \right) \left( 1+ \left( \frac{1+i}{2} \right)^2 \right) \left( 1+{\left( \frac{1+i}{2} \right)^2}^2 \right) \ldots \\
&\left( 1+{\left( \frac{1+i}{2} \right)^2}^k \right) \ldots \left( 1+{\left( \frac{1+i}{2} \right)^2}^{2001} \right) \text{.}
\end{align*}
Can someone help?Whatever I trying i can't find some rule to calculate this.
 A: Multiply by $$\left(1-\frac{1+i}{2}\right)$$ compute, and divide by that number at the end.
The thing is that $$\left(1-\frac{1+i}{2}\right)\left(1+\frac{1+i}{2}\right)=\left(1-\left(\frac{1+i}{2}\right)^2\right)$$ then $$\left(1-\left(\frac{1+i}{2}\right)^2\right)\left(1+\left(\frac{1+i}{2}\right)^2\right)=\left(1-\left(\frac{1+i}{2}\right)^4\right)$$
and so on.
The whole product collapses to $$1-\left(\frac{1+i}{2}\right)^{2^{2002}}$$
Then we divide by $1-\frac{1+i}{2}$ to get back the original product: 
$$\begin{align}\frac{1-\left(\frac{1+i}{2}\right)^{2^{2002}}}{1-\frac{1+i}{2}}&=2\left(1-\left(\frac{1+i}{2}\right)^{2^{2002}}\right)\cdot\left(1-\frac{1-i}{2}\right)\\&=2\left[1-\left(\frac{1+i}{2}\right)^{2^{2002}}-\left(\frac{1+i}{2}\right)+\left(\frac{1+i}{2}\right)^{2^{2003}}\right]=2\left[1-\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2002}}-\left(\frac{1+i}{2}\right)+\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2003}}\right]\\&=2\left[1-\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2002}}-\left(\frac{1+i}{2}\right)+\left(2^{-1/2}e^{\pi i/4}\right)^{2^{2003}}\right]\\&=2\left[1-2^{-2^{2000}}-\frac{1+i}{2}+2^{-2^{2001}}\right]\\&=\left[\frac{3}{2}-2^{1-2^{2000}}+2^{-2^{2001}}\right]-i\end{align}$$
