Start with some notations: $(a,q)_n=(1-a)(1-aq)\cdots(1-aq^{n-1})$, shortened by $(a)_n$, and $(a)_{\infty}=\prod_{k=0}^{\infty}(1-aq^k)$.

It's easy to verify (using algebraic means) that, for each $m\in\mathbb{Z_{\geq0}}$, $$\sum_{n\geq0}\frac{q^{n^2}q^{mn}}{(q)_n(q)_{n+m}}=\frac1{(q)_{\infty}}. \tag1$$

QUESTION.What is a combinatorial (conceptual) reason for (1) to be independent of $m$?