TITLE: Partitions-sum of divisors identity
QUESTION [13 upvotes]: A few years ago I first read about the marvelous Euler identity:
$\sum_{n\in\mathbb{N}}p(n)z^n=\prod_{k\geq1}\frac{1}{1-z^k}$,
where $p(n)$ is the number of partitions of $n$ ($p(0)=1$ by convention) and some of its beautiful consequences (like the pentagonal number theorem). Taking log of both sides of Euler identity and differentiating, the following nice recursive formula magically appears:
$np(n)=\sum_{k=0}^{n-1}p(k)\sigma(n-k)$,
where $\sigma(n)$ denotes the sum of the divisors of $n$. After some googling I found this identity quoted in a few places, but always without any reference. Since I am quite ignorant about the theory of partitions and related matters, I would like very much to know:
1) Who discovered this identity? Does it have a name?
and the much more interesting: 
2) Is there a proof without generating functions?
Thank you!

REPLY [5 votes]: Part 1.) R. Stanley mentions on page 59 (in the answer to exercise 24a, which is the identity mentioned in the OP) in Enumerative Combinatorics vol. 1 (exercise is 78a in later edition) the following somewhat inconclusive fact:

Some related results are due to Euler and recounted in $\S$303 of P.A. MacMahon, Combinatory Analysis, vol. 2, Cambridge University Press, 1916...

Searching the Euler archive showed a few similar results (but those were on the generating function of $\sigma(n)$ being related to the logarithmic derivative of the Euler generating function.