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Cauchy's identity states that $$ \prod_{i\geq 0}\frac{1-axq^i}{1-xq^i}=\sum_{n\geq 0}\frac{(1-a)(1-aq)\cdots(1-aq^{n-1})}{(1-q)(1-q^2)\cdots(1-q^n)}x^n. $$

Is it possible to somehow derive this identity as a special case of the $q$-binomial theorem? Mathworld references that it follows as a special case, and I thought maybe setting $a=q^k$ for some power $k$ might lead to it, but I can't say for sure.

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    It looks suspiciously like one of the things Ramanujan was playing with...2011-12-15
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    The q-binomial theorem is the identity you wrote down above. So what do you mean by Cauchy's identity?2011-12-15

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