$\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)...$ How can I prove the identity $\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$ for $|x|<1$? I am preferably looking for a derivation rather than using the RHS. I have tried using binomial expansion, but it only seems to give the LHS back. I also tried taking the logarithm of $\frac1{1-x}$ on seeing a product and using the Taylor series of $\ln{(1+x)}$, but this appears to be a dead end.
 A: Hint: use the telescoping product$$\prod_{j\ge0}\left(1+x^{2^j}\right)=\prod_j\frac{1-x^{2^{j+1}}}{1-x^{2^j}}.$$
A: Use $(P+Q)((P-Q)=P^2-Q^2$, repeatedly:
$$F=(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)......=(1-x^2)(1+x^2)(1+x^4)(1+x^8).....=(1-x^4)(1+x^4)(1+x^8)(1+x^{16})....=(1-x^8)(1+x^8)(1+x^{16})...(1+x^{2^n})=(1-x^{2^{n+1}})$$
When $|x|<1$ and $n \to \infty$, then $F =1$,
$$\implies \frac{1}{1-x}=(1+x)(1+x^2)(1+x^4)(1+x^8)....., |x|<1.$$
A: First prove by induction that
$$(1+x)(1+x^2)\ldots(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x}$$
and then take the limit as $n\to\infty$.
A: I provide a combinatoric approach. Consider the generating function of how many ways we can express each non-negative integer using binary digits. Firstly note that this representation is unique and hence the generating function is just
$$1+x+x^2+x^3+\dots=\frac1{1-x}$$
Alternatively the generating function is the product
$$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})\cdots$$
because this encapsulates the fact that each digit can either be a zero and hence the associated product should just contain $1$ or the digit is one and hence the product should contain $x^{2^n}$ for the $n$th digit.
A: We can repeatedly double the power in the denominator as follows: (similar to rationalizing the denominator, if you are familiar wtih that)
$$
\begin{align}
&\frac1{1-x}=\frac1{1-x}\cdot\frac{1+x}{1+x}\\
=&\frac{1+x}{1-x^2}=\frac{1+x}{1-x^2}\cdot\frac{1+x^2}{1+x^2}\\
=&\frac{(1+x)(1+x^2)}{1-x^4}=\frac{(1+x)(1+x^2)}{1-x^4}\cdot\frac{1+x^4}{1+x^4}\\
=&\frac{(1+x)(1+x^2)(1+x^4)}{1-x^8}=\cdots
\end{align}
$$
Hopefully it is clear how this pattern continues forever. Now, since
$$
\lim_{n\to\infty}x^n=0
$$
when $|x| < 1$, we can see that the denominator approaches $1$. Therefore, the numerator is all that remains, and the identity is shown.
