Showing infinite product identity I'm having difficulty showing the following identity (on open unit disk):
$$\frac{1}{1-z} = \prod_{n=0}^\infty{1+z^{2^n}}$$
My first idea was to show that they agreed on a set with non isolated points and show that the product converges to an analytic function on the unit disk, so that by uniqueness, they are equal. However, aside from 0, I cannot think of more appropriate values to choose. As well, I'm not quite sure how to show the uniform convergence of the infinite sum of |z|^(2^n) on compact subsets of the unit disk (or uniform convergence on unit disk?) to be able to show that the product converges to an analytic function.
Thank you!
 A: Hint:
$$\frac{1}{1-z}=\prod_{n=0}^{\infty}(1+z^{2^n})$$
Recall that 
$$\displaystyle \frac{1}{1-z}=\sum_{n=0}^{\infty}z^{n}$$
So, the equality
$$\prod_{n=0}^{\infty}(1+z^{2^n})=\frac{1}{1-z}=\sum_{n=0}^{\infty}z^{n}$$
boils down to the statement:

Every even natural number (including $0$) can be written uniquely as the sum of distinct powers of $2$.

Try proving this by induction.
You will then have shown that these two objects are formally equal, and since they both converge on the disc, they are equal as functions. 
EDIT: I was being silly--see Joriki's comment below.
A: Here's another take on this. The basic tools are the same what are in the comments of the answer above but also we will exploit 
$$\prod_{n=0}^\infty(1+x_m)=\sum_{S\in\mathcal{P}_f(\mathbb{W})} \prod_{m\in S}{x_m}$$  
Which seems true but I don't know the name of this supposed identity. I have asked about using this identity here. 
Applying this to $\prod_{n=0}^\infty{1+z^{2^n}}$ we find
$$\prod_{n=0}^\infty{1+z^{2^n}}=\sum_{S\in\mathcal{P}_f(\mathbb{W})}{\prod_{n\in S}{z^{2^n}}}=\sum_{S\in\mathcal{P}_f(\mathbb{W})} z^{\sum_{s\in S} 2^{s_i}}=\sum_{n=0}^\infty {z^n}$$
The first equality is the identity we exploit. The second is just unpacking some of what a product is: we are using $z^az^b=z^{a+b}$. The last equality is the one of interest: It really holds because $g: \mathcal{P}_f(\mathbb{W})\to \mathbb{W}$ defined as $$g(\{s_1, \dots, s_k \})= \sum_{i}^k {2^{s_i}}$$ is in fact a bijection. Note that $g(\{\})=0$ and $g(\{0\})=1$. Indeed we are just formally saying "Every natural number can be written uniquely as the sum of 
 distinct (non-negative) powers of $2$. 
To finish the argument of the OP we will just say that the geometric sum identity is well known. And thereby we have established that 
$$\prod_{n=0}^\infty{1+z^{2^n}}=\sum_{n=0}^\infty {z^n}=\frac{1}{1-z}$$
And we somehow have avoided induction but this might be tucked into the argument which establishes the product sum identity above. I am not sure. 
