A combinatorial proof: where art thou?

Question

Start by introducing the finite sums $$A_n:=\sum_{m=1}^nq^m\prod_{j=1}^{m-1}(1-q^j) \qquad \text{and} \qquad B_n:=\sum_{m=1}^nq^m\prod_{j=m+1}^n(1-q^j).$$ An algebraic proof is facile: Clearly, $A_1=B_1=q$. Note $A_{n+1}=A_n+q^{n+1}\prod_{j=1}^n(1-q^j)$ while $B_{n+1}=B_n+q^{n+1}-q^{n+1}B_n$. By induction $A_n=B_n$, so it remains to verify $\prod_{j=1}^n(1-q^j)=1-B_n$ or $\sum_{\pmb{{\color{blue}{m=0}}}}^nq^m\prod_{j=m+1}^n(1-q^j)=1$ or $\sum_{m=0}^n\frac{q^m}{(q)_m}=\frac1{(q)_n}$. The latter is valid for $n=0$ and if true for $n$ then $\sum_{m=0}^{n+1}\frac{q^m}{(q)_m}=\frac{q^{n+1}}{(q)_{n+1}}+\sum_{m=0}^n\frac{q^m}{(q)_m}=\frac{q^{n+1}}{(q)_{n+1}}+\frac1{(q)_n}=\frac{q^{n+1}}{(1-q^{n+1})(q_n)}+\frac1{(q)_n}=\frac1{(q)_{n+1}}$. $\square$

It's worth remarking that $A_{\infty}=B_{\infty}=1+\sum_{k\in\mathbb{Z}}(-1)^{k-1}q^{\omega_k}$ where $\omega_k=\frac{k(3k+1)}2$ are the pentagonal numbers.

QUESTION. Can you furnish a direct combinatorial proof for $A_n=B_n$?

Answer 1

Consider all partitions $\lambda$ with distinct parts not exceeding $n$ and sum up $(-1)^{r(\lambda)+1}q^{|\lambda|}$, where $r$ goes for the number of parts. You may count the sum by fixing the smallest part, or by fixing the largest part, getting representations $B_n, A_n$ respectively.