mod 5 partition identity proof

Question

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:

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

Answer 1

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.

Answer 2

It is not about five at all, as may be seen from Julian Rosen's answer, but let me say it in more explicit form.

Simply denote $x^5=y$ and get more general identity $$\prod_{m=0}^\infty \frac{1}{1-xy^m}=\sum_{i=0}^\infty \frac{x^i}{(1-y)(1-y^2)\dots (1-y^i)}.\,\,\,(*)$$ Coefficient of $x^a y^b$:

on the left - is the number of partitions of $b$ into at most $a$ parts;

on the right - is the number of partitions of $b$ into parts not exceeding $a$.

Yes, it is the same by conjugation.

If you ask for purely algebraic solution (conjugation is rather combinatorial bijection), you may note that each part of $(*)$ satisfies initial condition $F(0,0)=1$ and the functional equation $F(x,y)(1-x)=F(xy,y)$, which determine unique formal power series.