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.