Proof $\prod_{i=0}^{\infty}(1+x^{2^i})=\frac{1}{1-x}$ both algebraically and combinatorially

Question

I was doing some school work review and came across this equation $\prod_{i=0}^{\infty}(1+x^{2^i})=\frac{1}{1-x}$.

I'm suppose to prove this combinatorially but I was having trouble to believe that this equation holds in the first place. So can someone enlighten me with its proof both combinatorially and algebraically?

I know that the RHS expands into $\sum_{i=0}^{\infty}x^i$ but how does the LHS equal to the RHS? If I attempt to expand the LHS I get something like $x^2+x^4+x^6+...$ with varing coefficients.

Thanks!

Edit: As multiple people suggested, the original equation should be $\prod_{i=0}^{\infty}(1+x^{{2}^{i}})=\frac{1}{1-x}$ instead, otherwise the equation does not hold.

Answer 1

The equation should be:

$$\prod_{i=0}^{\infty}(1+x^{2^i})=\frac{1}{1-x}$$

which makes sense combinatorially because every integer has a unique binary representation, and the algebraic expansion of the LHS follows the same pattern.

Answer 2

Hint :- $$1 + x + x^2 + x^3+x^4+x^5...= (1+x)(1+x^2+x^4...) = (1+x)(1+x^2)(1+x^4+x^8...)$$

You can continue this indefinitely to regenerate the LHS