Euler's Bernoulli number Identity help

Question

My professor and I derived the $n$th Bernoulli number as the recursion $$B_n=\frac{-1}{n+1}\sum_{k=0}^{n-1}\binom{n+1}{k}B_k.$$ Later, in a paper I was reading, there is a similar identity attributed to Euler, where $$-(n+1)B_n=\sum_{k=2}^{n-2}\binom{n}{k}B_kB_{n-k}$$

Certainly reindexing the first equation would help to get $$-(n-1)B_n=\sum_{k=0}^{n-2}\binom{n}{k}B_k$$

But now the problem lies with the multiplication of each summand by the $(n-k)$th Bernoulli number and this, along with the index starting at 2 gives the identity. I'm failing to see the math in this. The paper I'm reading states that $b(x)=\frac{x}{e^x-1}=\sum_{k=0}^\infty\frac{B_nx^n}{n!}$ yields the result, but I'm not seeing how. Is there an online proof I can read to help clarify?

EDIT: The paper I'm reading actually states that the Euler Identity is a consequence of $$b(x)^2=(1-x)b(x)-xb'(x)$$ This is really where I'm seeing confusion. I've proved the first identity independently using the ideas provided in the answer below, but not being familiar with Laurent series I'm having trouble. I'm hoping, with the new identity below that someone can offer a little more insight. Thank you.

Answer 1

To prove the first identity it is sufficient to consider that $$ \frac{e^x-1}{x}\cdot\frac{x}{e^x-1}=1.$$ Now write both factors as Taylor series around $x=0$ and take the Cauchy product of such series.

To prove Euler's identity, consider that: $$\frac{d}{dx}\frac{1}{e^x-1} = -\frac{e^x}{(1-e^x)^2} = \frac{1}{1-e^x}\cdot\left(-1+\frac{1}{1-e^x}\right)$$ then take the Laurent series of the LHS around $x=0$ and compare it with the product of the Laurent series for the factors in the RHS.

Answer 2

I think I figured it out.

So I have $$\sum_{j=0}^{k}{\binom{k}{j}B_jB_{k-j}}=(1-k)B_k-kB_{k-1}$$ But $$\sum_{j=0}^{k}{\binom{k}{j}B_jB_{k-j}}=2B_0B_k+2B_1B_{k-1}+\sum_{j=2}^{k-2}B_jB_{k-j}$$ Since $B_0=1, B_1=-1/2, \binom{k}{0}=1, \binom{k}{1}=k$, $$\sum_{j=0}^{k}{\binom{k}{j}B_jB_{k-j}}=2B_k-kB_{k-1}+\sum_{j=2}^{k-2}B_jB_{k-j}=B_k-kB_k-kB_{k-1}$$ Therefore, the unneeded $B_{k-1}$ term disappears and I'm left with $$\sum_{j=2}^{k-2}\binom{k}{j}B_jB_{k-j}=-B_k-kB_k=-(k+1)B_k$$