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.