TITLE: How to prove it? (one of the Rogers-Ramanujan identities)
QUESTION [8 upvotes]: Prove the following identity (one of the Rogers-Ramanujan identities) on formal power series by interpreting each side as a generating function for partitions:
$$1+\sum_{k\geq1}\frac{z^k}{(1-z)(1-z^2)\cdots(1-z^k)}=\prod_{k\geq1}\frac{1}{1-z^k}$$

REPLY [3 votes]: Let us use the notation $[z^n] \sum_k a_k z^k$ to represent the operation of extracting $a_n$, the coefficients of $z^n$, from a formal power series $\sum_k a_k z^k$.
Let us look at the components in RHS of the identities we want to show:
$$1+\sum_{k\geq1}\frac{z^k}{(1-z)(1-z^2)\cdots(1-z^k)}=\prod_{k\geq1}\frac{1}{1-z^k}\tag{*}$$
We know:
$$[z^n]\frac{1}{1 - z^k} = \begin{cases}1,&k|n\\ 0,&\text{otherwise}\end{cases}$$
This represent the tautology that the number of ways to represent $n$ as a multiple of $k$ is $1$ if $k|n$ and $0$ otherwise.
Similarly, the expression:
$$[z^n]\left(\frac{1}{1-z^k}\frac{1}{1-z^l}\right)$$
is the number of ways of breaking $n$ into 2 unordered pieces, one piece is a multiple of $k$ while the other is a multiple of $l$. In general,
$$[z^n] \prod_{k\ge 1}\frac{1}{1-z^k}$$
is the number of ways of breaking $n$ into sum of integers $\ge 1$. Once again, order of the breakdown doesn't matter.
In the LHS of $(*)$, the $k$-th term:
$$[z^n]\frac{z^k}{(1-z)(1-z^2)\cdots(1-z^{k})}$$
is the number of ways to represent $n$ as sum of positive integers $\le k$ and the factor $z^k$ in numerator constraints the sum to use the integer $k$ at least once. ie. if $n = c_0 + c_1 + \cdots + c_r$, is a breakdown of $n$ as sum of integers $c_s \ge 1$, the $k$ corresponds to 
$\max( \{ c_1, \ldots c_r \} )$, the maximum of integer appear in a breakdown.
The $\sum_{k}$ part in LHS of $(*)$ can now be interpreted as a sub sum of RHS with respect to the maximum of integer appear in a breakdown. The extra $1$ in LHS is used to cover the special case in RHS where $0$ is treated as a sum of zero number of positive integers.