Euler's theorem by generating functions

Fix $N\ge 1$. Using $1+x^m=(1-x^{2m})(1-x^m{)}^{-1}$ for $1\le m\le N$ and canceling only the even factors that already appear in the denominator gives ${\prod }_{m=1}^N(1+x^m)=({\prod }_{m=\lfloor N\mathrm{/}2\rfloor +1}^N(1-x^{2m}))/({\prod }_{\begin{array}{c}1\le j\le N\\ j\text{ odd}\end{array}}(1-x^j))$. Every uncancelled numerator factor has degree $>N$, so the left-hand side and $1\mathrm{/}{\prod }_{1\le j\le N,j\text{ odd}}(1-x^j)$ have the same coefficients through degree $N$.

Fix $n\ge 0$ and choose $N>n$. By step 1.1, the coefficient of $x^n$ in ${\prod }_{m=1}^N(1+x^m)$ equals the coefficient of $x^n$ in $1\mathrm{/}{\prod }_{1\le j\le N,j\text{ odd}}(1-x^j)$. Factors with index $>N$ have degree $>n$ in both products, so this is also the common coefficient of $x^n$ in the two infinite products ${\prod }_{m\ge 1}(1+x^m)$ and ${\prod }_{m\ge 1}(1-x^{2m-1}{)}^{-1}$.

Substitute the two product expansions from [L1] and [L2] into step 2.1. The resulting formal series are equal, so [L3] gives equality of every coefficient. Hence $p_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}}(n)=p_{\mathrm{o}\mathrm{d}\mathrm{d}}(n)$ for all $n\ge 0$.