Conjectured analogue of Fermat's Little Theorem for Bernouli numbers
Solution 1:
If you work out the powers of a prime dividing $D_{2n}$ you get the following which is a well-known stronger version of the Clausen-von Staudt result.
If $p=2$, $\nu_2(D_{2n})=\nu_2(n)+3$. If $p$ is odd and $(p-1)\mid 2n$ then $\nu_p(D_{2n})=\nu_p(n)+1$, otherwise $\nu_p(D_{2n})=0$.
Now for the congruence to hold you just need to check it modulo these powers of each prime and this is easy because of the structure of the group of units module a prime power.