Eulerian number identity The Eulerian number $A(n,m)$ is defined as the number of permutations $\sigma \in S_n$ having precisely $m$ descents, i.e. indices $i$ such that $\sigma(i)>\sigma(i+1)$. 
The wikipedia entry on these numbers mentions one identity for which I cannot locate a proof, and I would be grateful for a reference. Namely
$$\sum_{m=0}^{n-1} (-1)^m \frac{A(n,m)}{\left(n\atop m\right)}=(n+1) B_n\ \ \forall n\ge 2.$$ Here $\{B_n\}$ are the Bernoulli numbers.
(I do understand the two related identities immediately preceding this one. The first of these can be found in many places, and the second is easy.)
 A: Here's a proof of the identity. Unfortunately, it doesn't show how anyone would come up with this formula. Like the proofs referred to in the comments, this proof is based on the beta integral 
$$\int_0^1 u^m (1-u)^{n-m}\, du = \frac{m!\,(n-m)!}{(n+1)!} = \frac{1}{(n+1)\binom{n}{m}}.$$
By the generating function for the Eulerian numbers, which can be found in the Wikipedia article (https://en.wikipedia.org/wiki/Eulerian_number), we have
$$
1+\sum_{n=1}^\infty \frac{x^n}{n!} \sum_{m=0}^{n-1} A(n,m) t^m = \frac{t-1}{t-e^{(t-1)x}}.
$$
Replacing $t$ with $-u/(1-u)$ and $x$ with $x(1-u)$, and simplifying, gives
$$
1+\sum_{n=1}^\infty \frac{x^n}{n!} \sum_{m=0}^{n-1} (-1)^m A(n,m) u^m (1-u)^{n-m} = \frac{e^x}{1+(e^x-1)u}.
$$
Integrating with respect to $u$ from 0 to 1 gives
$$
1+\sum_{n=1}^\infty \frac{x^n}{n!} \sum_{m=0}^{n-1} (-1)^m 
\frac{A(n,m)}{(n+1) \binom nm}=e^x\frac{x}{e^x-1} =1+\frac{x}{2}+\sum_{n=2}^\infty B_n \frac{x^n}{n!}.
$$
A: In the references section in the article of wikipedia, an article of Worpitzky is cited. The identity you are looking for is the first (unnumbered) equation in page 222. (One must be careful since Worpitzky uses "old" definitions of the numbers involved.)
