This first request may be easy, but the asymptotics for the next step has me scratching my head.
Through an informal inductive argument I have been able to show
$$ (1) \quad \sum_{j=0}^{n-1}2^{2m(n-j)}E_{2j}\binom{2n}{2j} = 2\sum_{j=1}^{2^{m-1}}(-1)^j(2j-1)^{2n} , \quad m \ge 2.$$
The $E_{2j}$ are the Euler numbers. In trying to extend this to integers greater than two and I have acquired numerical evidence for
$$ (2) \quad \sum_{j=0}^{n-1}r^{2m(n-j)}E_{2j}\binom{2n}{2j} \sim 2 (-1)^r\sum_{j=1}^{r^{m-1}}(-1)^j(rj-1)^{2n} , \quad m,r=2,3,...$$
I have searched for an appropriate identity but have only been able to postulate an asymptotic relationship. Note that because of the factor of $(-1)^r$ and the $r$ in the upper limit of the summation it is likely that (2) applies only to integer $r>1.$ A different proof than mine might help in my own investigations.
My question is: can (2) be proven, and if that is not possible, can (1) be proven without an induction; e.g., generating functions.