Factorisation and roots of infinite polynomial Recently, I’ve learned about generating functions. From my understanding you basically represent each “option” with a polynomial, and the resulting polynomials multiplied would give coefficients related to the number of ways you can choose some objects.
So here, I’m trying to apply the same thing to binary numbers! Using $x^{2^n}+1$ to represent 2^n in binary and multiplying first 3 terms of it i.e. $(x+1)(x^2+1)(x^4+1)=\sum_{i=0}^7x^i$ I can show that 0-7 has unique representation in binary!  
However, I would like to prove this for all natural numbers. Simple, simply multiply $\Pi_{i=0}^\infty (x^{2^i}+1) $ what I would get would be $1+x+x^2+...$. Ayyy! $\frac{1}{1-x}$! I quickly write down
Some might see the problem here, but we said that $\frac{1}{1-x}=\Pi_{i=0}^\infty x^{2^i}+1$. The problem here is that (I GIVE UP PHONE LATEX) LHS has no roots while RHS has infinitely many roots i.e. x=1, x^2=-1, ...
$$\frac{1}{1-x}=\Pi_{i=0}^\infty x^{2^i}+1 $$
question: why do they not match in terms of roots, even though they “clearly is the same”?
Also I believe you can do the same “factorisation” in different “base”. Is the factorisation even valid? It’s not unique though... 
Gareth 
Sorry it’s hard to type and format latex on phone

They are not the same. $\frac{1}{1 - x} = \sum_{i = 0}^\infty x^i$ only when $|x| < 1$. All the roots of the infinite polynomial occur when this condition is not satisfied. In fact, $|r| = 1$ for all roots $r$.
You can do this factorization in any base if you include as the exponents all multiples of the powers of the base up to $b - 1$. So for base three, the product would be $\prod_{i = 0}^\infty (1 + x^{3^i}+x^{2 \cdot 3^i})$
