$Z_0=\left(\frac{1-i}{2}\right)$  then the value of the product $\left(1+z_0\right){\left(1+{z_0}^2\right)}^{ }(1+z_0^{2^2})\mathrm{.....}(1+z_0^{2^n})$  must be

Let $p=\left(1+z_0\right){\left(1+{z_0}^2\right)}^{ }(1+z_0^{2^2})\mathrm{.....}(1+z_0^{2^n})$

$(1-z_0)P={(1-{z_0}^2)}^{ }{(1+z_0^2)}^{ }(1+z_0^{2^2})\mathrm{.....}(1+z_0^{2^n})$ $=1-z_0^{2^{n+1}}$

$\Rightarrow P=\frac{[1-z_0^{2^{n+1}}]}{(1-z_0)}$

$We have Z_0^2=(-\frac{i}{2})$

$Z_0^{2^{n+1}}={(-\frac{i}{2})}^{2^n}=\frac{{(-1)}^{2^n.}{\left(i\right)}^{2^n}}{2^{2^n}}$  

         $=\frac{1}{2^{2^n}}$

$P=\frac{[1-z_0^{2^{n+1}}]}{(1-z_0)}$

   $=\frac{(1-\frac{1}{2^{{ }^{2^n}}})}{(1+i)/2}=\frac{2}{(1+i)}[1-\frac{1}{2^{2^n}}]$