$[1+(\frac{1+i}{2})][1+(\frac{1+i}{2})^2][1+(\frac{1+i}{2})^{2^2}]...[1+(\frac{1+i}{2})^{2^n}]=(1-\frac{1}{2^{2^n}})(1+i)$,where $n\ge 2$ 
Show that $$\!\!\!\left[1+\left(\frac{1+i}{2}\right)\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^2\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^{2^2}\right]\cdots\left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]\!\!\!=\left(1-\frac{1}{2^{2^n}}\right)(1+i)$$ for $n\ge 2$.


I took $\frac{1+i}{2}=\frac{1}{\sqrt2}e^{i\frac{\pi}{4}}$ and tried solving but i could not reach the RHS.Please help.
 A: Hints:


*

*$(1-a) \cdot (1+a) = 1 - a^2\,$, $\,(1-a) \cdot (1+a)(1+a^2) = 1 - a^4\,$, $\;\ldots$

*$(1+i)^2 = 2i$

*$1 - \dfrac{1+i}{2}=\dfrac{1-i}{2} = \dfrac{1}{1+i}$
A: Notice that, for any $x \neq 1,$
\begin{align*}
\prod_{k=0}^n(1+x^{2^k})&=(1+x)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})\\&=\frac{(1-x)(1+x)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})}{1-x}\\&=\frac{(1-x^2)(1+x^2)(1+x^{2^2})\cdots(1+x^{2^n})}{1-x}\\&\vdots\\&=\frac{1-x^{2^{n+1}}}{1-x}.
\end{align*}
Thus, \begin{align*}\prod_{k=0}^n\left[1+\left(\frac{1+i}{2}\right)^{2^k}\right]&=\frac{1-\left(\dfrac{1+i}{2}\right)^{2^{n+1}}}{1-\dfrac{1+i}{2}}=\frac{1-\left[\left(\dfrac{1+i}{2}\right)^2\right]^{2^{n}}}{\dfrac{1}{1+i}}\\&=\left[1-\left(\frac{i}{2}\right)^{2^n}\right](1+i)\\&=\left(1-\frac{1}{2^{2^n}}\right)(1+i).\end{align*}
A: Hint: Use the general case
$$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})=\dfrac{(1-x^{2^{n+1}})}{1-x}$$
A: Hint: Use mathematical induction. Also, for $n > 2$,  $$\left(\frac{1+i}{2}\right)^{2^{n}} = \frac{1}{2^{2^{n-1}}},$$
and
$$\left(1-\frac{1}{2^{2^{n-1}}}\right)\left(1+\frac{1}{2^{2^{n-1}}}\right)=1-\frac{1}{2^{2^{n}}}.$$
