Euler could run circles around this one.

One way to find the answer is by analyzing and simplifying the expression. We can convert the product to a sum using the property of exponents that $a^b \cdot a^c = a^{b+c}$ :

$\large \displaystyle \prod_{k=1}^{\infty} e^{(i \pi 2^{-k})} = \large e^{\left(\displaystyle \sum_{k=1}^{\infty} (i \pi 2^{-k})\right)}$

Simplifying the sum, we get

$\large e^{\left(i \pi \displaystyle \sum_{k=1}^{\infty} 2^{-k}\right)}$

That sum is familiar to anyone who has considered Zeno's paradox (https://en.wikipedia.org/wiki/Zeno%27s paradoxes#Dichotomy paradox), but it can be evaluated using the theorem $\displaystyle \sum_{k=1}^{\infty} r^k = \frac{r}{1-r}, 0<r<1$ . In this case, $r=1/2$ .

$\large \displaystyle e^{i \pi \frac{1/2}{1-1/2}}$

$\large \displaystyle e^{i \pi}$

This is part of Euler's identity, and it is equal to $-1$ .

There is another way of solving this that is more visually appealing. In the complex plane, multiplying by a factor of $e^{i\alpha}$ can be interpreted as a rotation of a number about the origin by an angle of $\alpha$ . Therefore, each term in the product is a rotation by $\pi 2^{-k}$ . If start at $1$ , and rotate first by $\pi / 2$ , it takes us to a position of $i$ . Rotating then by $\pi / 4$ takes us halfway between the positive imaginary axis and the negative real axis. Each rotation takes us closer and closer to a position of $-1$ . This process of repeated rotations can be seen in the animation below.

$\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 8 } +...=1$ , so we end up with Euler's equation ${e}^{i \pi}=-1$