Any hint for this product of complex numbers I have to calculate the modulus of $z$. I've already tried to find a general formula for $\left(\frac{1+i}{2}\right)^{2^n}$, which seems to be $\frac{1}{2^{2^{n-1}}}, \forall \geq3$, using the trigonometric form and de Moivre's formula.

$z=\left[1+\frac{1+i}{2}\right] \left[1+\left(\frac{1+i}{2}\right)^2\right] \left[1+\left(\frac{1+i}{2}\right)^4\right] \cdots \left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]$

How should I keep solving this?

Thanks!
Note that 

$$z=
\frac {[1-(\frac {1+i}{2})]\left[1+\frac{1+i}{2}\right] \left[1+\left(\frac{1+i}{2}\right)^2\right] \left[1+\left(\frac{1+i}{2}\right)^4\right] \cdots \left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]        }{[1-(\frac {1+i}{2})]}=$$

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

We may simplify it with $$(\frac{1+i}{2})^2 = \frac {i}{2}$$
Use $\frac {1+i} 2=\frac 1 {\sqrt 2} e^{i\pi/4}$. Since $|e^{i\theta}|=1$ for all real $\theta$ we get  $|\frac {1+i} 2|^{2^{n}}=\frac 1 {2^{2^{n-1}}}$.