Is there any proof of this identity? 
$$\prod\limits_{i=1}^{\infty}\frac{1}{1-yx^{i}}=\sum\limits_{k=0}^{\infty}\frac{y^k x^k}{(1-x)(1-x^2)...(1-x^k)}$$ 

I know its combinatorial proof through "inspection",but is it true when $x,y$ are not integers?are there any non-combinactorial proof of this indentity? 
 A: That is an equality of generating functions.  The number of partitions of $n$, $p(n)$, is given by the number of ways for writing $n$ as the sum of some $1s$, some $2s$, some $3s$... In other terms,
$$p(n)= [z^n](1+z+z^2+z^3+\ldots)(1+z^2+z^4+\ldots)(1+z^3+z^6+\ldots)\cdots $$
or, by setting $p(0)=1$,
$$ \sum_{n\geq 0} p(n)\,z^n=\frac{1}{1-z}\cdot\frac{1}{1-z^2}\cdot\frac{1}{1-z^3}\cdot\ldots = \prod_{h\geq 1}\frac{1}{1-z^h}. $$
If we introduce a tag-variable $y$, the coefficient of $x^n$ in $\prod_{i\geq 1}\frac{1}{1-y x^i}$ is a polynomial in the $y$-variable whose evaluation at $y=1$ returns $p(n)$. The coefficient of $y^a x^n$ in such product is the number of partitions of $n$ with exactly $a$ parts. Of course, if a partition has $a$ parts it is related to a number $\geq a$. Given a partition with $a$ parts, by transposing its Ferrer diagram we get a partition with each part having cardinality $\leq a$ and at least one part with cardinality $=a$. This is enough to explain the equality between the shown generating functions.
