Prove that $\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$ Prove that
$$\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$$
Here I am trying the following
\begin{align*}
\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} &= \prod_{i\geq 1} \left( \sum_{n \geq 0} x^{n}y^{n(2i-1)} \right)\\ 
&= \sum_{m\geq 0} \sum_{n \geq 0} p(2m-1,n)x^{n}y^{2m-1}\\ 
&= \sum_{n\geq 0}\left( \sum_{m \geq 0} p(2m-1,n)y^{2m-1}\right)x^{n} \\ 
&= \sum_{n\geq 0} \frac{y^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}x^{n}\\ 
\end{align*}
However, I am not sure if this is totally correct.
 A: 
We obtain
\begin{align*}
\color{blue}{\prod_{k=1}^\infty\frac{1}{1-xy^{2k-1}}}&=\prod_{k=0}^\infty\frac{1}{1-xy^{2k+1}}\\
&=\prod_{k=0}^{\infty}\frac{1}{1-\left(xy\right)\left(y^{2}\right)^k}\\
&=\frac{1}{\left(xy;y^2\right)_{\infty}}\tag{1}\\
&=\sum_{n=0}^{\infty}\frac{1}{\left(y^2;y^2\right)_n}(xy)^n\tag{2}\\
&=\sum_{n=0}^\infty\frac{1}{\prod_{k=0}^{n-1}\left(1-\left(y^2\right)^{k+1}\right)}\left(xy\right)^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\frac{1}{\prod_{k=1}^{n}\left(1-y^{2k}\right)}\left(xy\right)^n}\\
\end{align*}

Comment:

*

*In (1) we use the $q$-Pochhammer symbol
\begin{align*}
(a;q)_{n}&=\prod_{k=0}^{n-1}\left(1-aq^k\right)\\
(a;q)_{\infty}&=\prod_{k=0}^{\infty}\left(1-aq^k\right)\qquad\qquad |a|<1,|q|<1
\end{align*}


*In (2) we use the identity
\begin{align*}
\color{blue}{\frac{(at;q)_{\infty}}{(t;q)_{\infty}}=\sum_{n=0}^\infty\frac{(a;q)_n}{(q;q)_n}t^n\qquad\qquad |q|<1,|t|<1}\tag{3}
\end{align*}
where we set $a=0$. This is Theorem 2.1 in The Theory of Partitions by G. E. Andrews. We follow the proof there and write (3) as
\begin{align*}
\prod_{n=0}^\infty\frac{1-atq^n}{1-tq^n}=\sum_{n=0}^\infty\prod_{k=0}^{n-1}\frac{1-aq^k}{1-q^{k+1}}t^n
\end{align*}
The left-hand side is a function $F(t)$ which we want to expand as generating function
\begin{align*}
F(t)=\prod_{n=0}^\infty \frac{1-atq^n}{1-tq^n}=\sum_{n=0}^\infty A_n t^n\tag{4}
\end{align*}
in the unknown $A_n=A_n(a,q)$. Multiplying (4) with $1-t$  gives
\begin{align*}
(1-t)F(t)&=(1-at)\prod_{n=1}^\infty \frac{1-atq^n}{1-tq^n}=(1-at)\prod_{n=0}^\infty\frac{1-atq^{n+1}}{1-tq^{n+1}}\\
&= (1-at)F(tq)\\
&=(1-at)\sum_{n=0}^\infty A_nt^nq^n\tag{5}
\end{align*}
Denoting with $[t^n]$ the coefficient of $t^n$ in a series and making coefficient comparison in (5) gives
\begin{align*}
[t^n](1-t)F(t)&=A_n-A_{n-1}\\
&=[t^n](1-at)F(tq)\\
&=q^nA_n-aq^{n-1}A_{n-1}
\end{align*}
and
\begin{align*}
A_n(1-q^n)&=A_{n-1}\left(1-aq^{n-1}\right)\tag{6}\\
A_0&=1
\end{align*}
follows. From the recurrence relation (6) we obtain for $n\geq 1$:
\begin{align*}
\color{blue}{A_n}&=\frac{1-aq^{n-1}}{1-q^n}A_{n-1}\\
&=\frac{\left(1-aq^{n-1}\right)\left(1-aq^{n-2}\right)}{\left(1-q^n\right)\left(1-q^{n-1}\right)}A_{n-2}\\
&=\prod_{k=0}^{n-1}\frac{1-aq^k}{1-q^{k+1}}A_0\\
&\,\,\color{blue}{=\frac{(a;q)_n}{(q;q)_n}}
\end{align*}
and the claim (3) follows.
A: The question asks

Prove that
$$\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$$

As a first simplifying step, replace $\,x\,$ with $\,x/y\,$ to get
$$ \prod_{i\geq 0}\frac{1}{1-x\,y^{2i}} = \sum_{n\geq 0} \frac{x^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}. \tag{1} $$
Next, replace $\,y^2\,$ with $\,q\,$ to get
$$ \prod_{i\geq 0}\frac{1}{1-x\,q^i} = \sum_{n\geq 0} \frac{x^{n}}{\prod_{i=1}^{n}\left( 1-q^i \right)}. \tag{2} $$
The Wikipedia article Q-Pochhammer symbol
section on Combinatorial interpretation states

The $q$-Pochhammer symbol is closely related to the enumerative combinatorics of partitions. The coefficient of $\,q^ma^n\,$ in
$$ (a;q)_\infty^{-1} = \prod_{k=0}^\infty (1-a\,q^k)^{-1} $$ is the
number of partitions of $\,m\,$ into at most $\,n\,$ parts.


Since, by conjugation of partitions, this is the same as the number of partitions of
$\,m\,$ into parts of size at most $\,n\,$, by identification of generating series we obtain the identity:
$$ (a;q)_\infty^{-1} = \sum_{k=0}^\infty \left( \prod_{j=1}^k \frac1{1-q^j}\right)a^k
= \sum_{k=0}^\infty \frac{a^k}{(q;q)_k} $$ as in the above section.

and this last identity is exactly equation $(2)$ above except that
$\,k\,$ is used instead of $\,n\,$ as the summation index,
$\,a\,$ is used instead of $\,x\,$ and it is written using the
$q$-Pochhammer symbol on both sides.
Here is a proof in the style of the attempt in the question.
Define a generating function
$$ L:=\prod_{i\geq 1}\frac{1}{1-x\,y^{2i-1}}. \tag{3} $$
Use the Taylor series of $\,\frac1{1-z}\,$ to get
$$ L= \prod_{i\geq 1} \left( \sum_{n \geq 0} x^{n}y^{n(2i-1)} \right).
\tag{4} $$
Define $\,q(m,n)\,$ to be the number of partitions of $\,m\,$ into exactly
$\,n\,$ odd parts to get
$$ L = \sum_{m\geq 0} \sum_{n \geq 0} q(m,n)\,x^{n}y^m. \tag{5} $$
Reverse the order of summation to get
$$ L = \sum_{n\geq 0}\left( \sum_{m \geq 0} q(m,n)\,y^m\right)x^{n}. \tag{6} $$
Use the fact that the number of partitions enumerated by $\,q(m,n)\,$
is the same as the number of partitions of $\,m-n\,$ into at most $\,n\,$
even parts, since if we subtract $1$ from each part
the odd parts become even parts and the $1$ parts become $0$ parts, to get
$$ L=\sum_{n\geq 0} \frac{y^{n}}{\prod_{i=1}^{n}\left(1-y^{2i} \right)}x^{n}. 
\tag{7} $$
Combine the $n$th powers to get the final result
$$ L = \sum_{n\geq 0} \frac{(x\,y)^{n}}{\prod_{i=1}^{n}\left(1-y^{2i} \right)}.
\tag{8} $$
A: One perspective is to simply write
$$p=\dfrac{1}{(1-xq)(1-xq^2)(1-xq^3)\cdots}=A_0+\dfrac{A_1}{1-xq}+\dfrac{A_2}{(1-xq)(1-xq^2)}+\cdots$$ and multiply each consecutive term while setting the numerator equal to $1$. This process, assuming $A_0=1$, gives
$$A_0(1-xq)+A_1=1\implies A_1=xq;$$ and that
$$p=\dfrac{1}{(1-xq)\cdots(1-xq^n)}+ \dfrac{A_{n+1}}{(1-xq)\cdots(1-xq^{n+1})}+\cdots$$ for all integers $n\geq 1$,  and therefore that $$A_n=xq^n.$$
While not a proof, this is a simple outline that can be easily made into one. Either this, or you can use the fact that
$$a_0\prod_{n=1}^{\infty}\left(1+\dfrac{a_n}{\sum_0^{n-1}a_k}\right)=\sum_{n=0}^{\infty}a_n,$$and put $a_n=\dfrac{xq^n}{(1-xq)\cdots (1-xq^n)}.$
