Throughout this question $n$ is a positive integer greater than 1. Consider the following well-known identity by Euler, $$\sum_{k=1}^{n-1} \binom{2n}{2k}B_{2k}B_{2n-2k}=-(2n+1)B_{2n}.$$ Rather unconventionally, we can also write the above identity as $$\sum_{k=1}^{n} \binom{2n}{2k}B_{2k}B_{2n-2k}=-2n B_{2n}.$$

I was wondering if the following identity is also known. $$\sum_{k=1}^{n} \binom{2n}{2k}B_{2k}B_{2n-2k}=\sum_{k=1}^{n} 2^{2k} \binom{2n}{2k}B_{2k}B_{2n-2k}.$$

This would give the following nice identity $$\sum_{k=1}^{n}(2^{2k}-1) \binom{2n}{2k}B_{2k}B_{2n-2k}=0.$$

My proof involves Newton’s Identities and simple manipulations of the equation derived.