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 $A\_n=B\_n$?
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A combinatorial proof: where art thou?
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.


