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Recall that $(a;\,q)_\infty$ is the $q$-Pochhammer symbol: $$(a;\,q)_\infty=\prod_{n=0}^\infty(1-a \, q^n).\tag1$$ Its important special case $(q;\,q)_\infty=\prod_{n=1}^\infty(1-q^n)$ is sometimes ...