I am looking for a proof that:


$$\prod\_\limits{m=0}^\infty \dfrac{1}{(1-x^{5m+1})}=\sum\_\limits{i=0}^\infty \dfrac{x^i}{\prod\_\limits{j=1}^i (1-x^{5j})}$$


The left hand side expands into:


$$\dfrac{1}{(1-x)(1-x^6)(1-x^{11})\dots}$$


which gives the number of partitions into parts $1\mod5$.


The right hand side expands into:


$$1+\dfrac{x}{(1-x^5)}+\dfrac{x^2}{(1-x^5)(1-x^{10})}+\dfrac{x^3}{(1-x^5)(1-x^{10})(1-x^{15})}+\dots$$.


The only proof I have is a fairly brute force method.


EDIT: I think we have:


$$\dfrac{1}{(1-x)(1-ax)(1-a^2x)\dots}\\=1+\dfrac{x}{(1-a)}+\dfrac{x^2}{(1-a)(1-a^2)}+\dfrac{x^3}{(1-a)(1-a^2)(1-a^3)}+\dots$$



mod 5 partition identity proof
More generally, there is an equality of power series in two variables
$$
\prod\_\limits{m=0}^\infty \dfrac{1}{(1-t x^{5m+1})}=\sum\_\limits{i=0}^\infty \dfrac{t^i x^i}{\prod\_\limits{j=1}^i (1-x^{5j})}.
$$
The coefficient of $t^i x^n$ on the left hand side is the number of partitions of $\frac{n-i}{5}$ into at most $i$ pieces (by convention the number of partitions of a non-integer in $0$). The coefficient of $t^i x^n$ on the right hand side is the number of partitions of $\frac{n-i}{5}$ into pieces of size at most $i$. Conjugation gives a bijection between the two kinds of partitions.


