How to prove that $\frac{1}{1-z}=\prod_{n=0}^\infty (1+z^{2^n})$?

Question

A simple pole can be written as $\displaystyle{\frac{c}{c-z}=\prod_{n=1}^\infty e^{\frac{1}{n}\left(\frac{z}{c}\right)^n}}$. How does one show that when $c=1$, $\displaystyle{\frac{1}{1-z}=\prod_{n=0}^\infty (1+z^{2^n})}$?

Answer 1

Hint:

$$\dfrac{1-z^{2^{n+1}}}{1-z}=\prod_{r=0}^n(1+z^{2^r})$$ for $z\ne1$

Now if $|z|<1,$ $$\lim_{r\to\infty}z^r=0$$

Here $r=?$