Euler's pentagonal number theorem by Franklin's involution

Let $\lambda =({\lambda }_1>\cdots >{\lambda }_{\mathrm{\ell }})$ be a nonempty partition into distinct parts. Write $s={\lambda }_{\mathrm{\ell }}$ for its smallest part, and let $t$ be the largest index such that ${\lambda }_i={\lambda }_1-i+1$ for every $1\le i\le t$. Thus the first $t$ rows form the maximal upper-right staircase. If $s\le t$ and $\lambda \mathrm{\ne }(2s-1,2s-2,\dots ,s)$, define $\alpha (\lambda )$ by deleting the last part $s$ and adding $1$ to each of the first $s$ parts. If $s>t$ and $\lambda \mathrm{\ne }(2t,2t-1,\dots ,t+1)$, define $\beta (\lambda )$ by subtracting $1$ from each of the first $t$ parts and adjoining a new last part $t$.

In the first case of step 1.1, the partition $\alpha (\lambda )$ is still distinct: the first $s$ parts remain strictly decreasing, the last changed part satisfies ${\lambda }_s+1>{\lambda }_{s+1}$ because those rows were consecutive, and deleting the old last part decreases the number of parts by $1$. The first $s$ parts of $\alpha (\lambda )$ are consecutive and its smallest part is larger than $s$, so $\alpha (\lambda )$ falls under the second construction with parameter $t=s$, and $\beta (\alpha (\lambda ))=\lambda$.

In the second case of step 1.1, the partition $\beta (\lambda )$ is still distinct: maximality of $t$ gives ${\lambda }_t-1>{\lambda }_{t+1}$, while the new last part $t$ is smaller than the previous smallest part because $s>t$. The first $t$ parts of $\beta (\lambda )$ are consecutive and its smallest part is exactly $t$, so $\beta (\lambda )$ falls under the first construction with parameter $s=t$, and $\alpha (\beta (\lambda ))=\lambda$. Thus steps 2.1 and 2.2 define a sign-reversing involution on all nonexceptional nonempty distinct partitions.

The empty partition contributes the constant term $1$. The only nonempty distinct partitions excluded from step 1.1 are the two staircase families $(2k-1,2k-2,\dots ,k)$ and $(2k,2k-1,\dots ,k+1)$. Their sizes are $k+(k+1)+\cdots +(2k-1)=k(3k-1)\mathrm{/}2$ and $(k+1)+(k+2)+\cdots +2k=k(3k+1)\mathrm{/}2$, and each has exactly $k$ parts, so each contributes the sign $(-1{)}^k$.

By step 2.2, all nonexceptional nonempty distinct partitions cancel in opposite-parity pairs. Step 2.3 leaves only the empty partition and the two exceptional staircase families, so [F2] gives ${\prod }_{m\ge 1}(1-x^m)=1+{\sum }_{k\ge 1}(-1{)}^k(x^{k(3k-1)\mathrm{/}2}+x^{k(3k+1)\mathrm{/}2})$, equivalently the two-sided sum over $r\in \mathbb{Z}$.