<s>[INST] $\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)...$ 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.
 [/INST] 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 $F =1$,
$$\implies \frac{1}{1-x}=(1+x)(1+x^2)(1+x^4)(1+x^8)....., |x|<1.$$
</s>