# Infinite product $(1+z)\prod_{n=1}^{+\infty}(1+z^{2^{n}})$

I have to show that if $|z| < 1$, $z \in \mathbb{C}$, $$(1+z)\prod_{n=1}^{+\infty}(1+z^{2^{n}})= \frac{1}{1-z}$$

I want to understand how to do this kind of exercises, any hint ?

Consider the finite product $$P(N) = \prod_{n=0}^N (1+z^{2^n}).$$ Then observe that \begin{align*} (1-z)P(N) &= (1-z)(1+z)(1+z^2)(1+z^4)\ldots(1+z^{2^N}) \\ &= (1-z^2)(1+z^2)(1+z^4)\ldots(1+z^{2^N}) \\ &= (1-z^4)(1+z^4) \ldots (1+z^{2^N}) \\ &= (1-z^8) \ldots (1+z^{2^N}) = \ldots \\ &= 1-z^{2^{N+1}}. \end{align*} So if $|z| < 1$, $\displaystyle \lim_{N \to \infty} (1-z) P(N) = \lim_{N \to \infty} 1 - z^{2^{N+1}} = 1$, and the result immediately follows.
The Laurent series expansion of the function $\frac{1}{1-z}$ is well known and has a lot of applications: $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ If you can identify the terms of the expansion in that expression of yours...