Problem: Define $\delta(X) = \frac{\sqrt{n}}{1+\sqrt{n}}\bar{X}_n + \frac{1}{2(1+\sqrt{n})}$
We assume $\bar{X}_n|\theta \sim Bin(\theta,n)$.
It is known that $\delta(X)$ is the Bayes estimator of $\theta$ with respect to the $Beta(\sqrt{n}/2,\sqrt{n}/2)$ prior for squared error loss. Since its risk is constant, it is also minimax.
The problem at hand is to show that
$$\delta'(X) = \frac{\sqrt{n}}{1+\sqrt{n}}\frac{1}{n}\sum_i X_i^k + \frac{1}{2(1+\sqrt{n})}$$
Is Bayes for $\theta^k$ with respect to the same prior.
Since it's squared error loss, the Bayes estimate is given by $E[\theta^k |X]$ and we know that $[\theta|X] \sim Beta(\sqrt{n}/2+X,\sqrt{n}+n-X)$.
However, the $k^{th}$ moment for a Beta distribution seems to be (according to Wikipedia) a nasty product of $k-1$ terms. So I assume there must be an easier way to show the above is the Bayes estimator of the $k^{th}$ moment.
Is there an easier way forward?