One of the nice identities about two-fold products of Bernoulli numbers is $$ \sum_{k=0}^{n} \dfrac{(1-2^{1-k})(1-2^{k-n+1})}{(n-k)! k!} B_k B_{n-k} = \dfrac{1-n}{n!} B_n $$ In some web resources it is credited for Gosper. But I could not find a proof of this identity.
Does anyone know the proof given by Gosper, and also alternative proofs if there are any?
Since Bernoulli numbers can be defined through $$ \sum_{n\geq 0}\frac{B_n}{n!} z^n = \frac{z}{e^z-1} \tag{1}$$ we also have $$ \sum_{n\geq 0}\frac{B_n(1-2^{1-n})}{n!} z^n = \frac{z}{e^z-1}-2\frac{z/2}{e^{z/2}-1}=-\frac{z}{2\sinh\frac{z}{2}} \tag{2}$$ and the given (convolution) identity can be read as $$ \left(-\frac{z}{2\sinh\frac{z}{2}}\right)^2 = \frac{z}{e^z-1}-z\cdot\frac{d}{dz}\left(\frac{z}{e^z-1}\right)\tag{3} $$ that is straightforward to check.