$\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)...$ 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.

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.
