How to find UMVUE of $\theta^k$ when $x_1, \ldots, x_n$ is a sample from Bernoulli$(\theta)$? Let $x_1, x_2, \ldots, x_n$ be a random sample from the Bernoulli ($\theta$).
The question is to find the UMVUE of $\theta^k$.
I know the $\sum_1^nx_i$ is the complete sufficient statistics for $\theta$.
Is  $\left(\frac{\sum_1^nx_i}{n}\right)^k$ the estimator or any other possible estimator?
Could someone just help me?

Having that
$$\theta^m=P\{ X_1=x_1,X_2=x_2,...,X_m=x_m\}$$
An unbiased estimator for $\theta^m$ is
$$T=
\begin{cases}
1,  & if \ \ X_1=X_2= \, ... \,=X_m =1 \\
0, & in \ other \ case
\end{cases}$$
But
$$\begin{align} E[T|S=s] & = P\{X_1=1,X_2=1,...,X_m=1|S=s\}=\frac{P\{X_1=1,X_2=1,...,X_m=1,S=s\}}{P\{S=s\}} = \\\\ & = \begin{cases}
0,  & if \ \ m>s \\
\frac{\theta^m\binom{n-m}{s-m}\theta^{s-m}(1-\theta)^{n-s}}{\binom{n}{s}\theta^s(1-\theta)^{n-s}}, & if \ \ m\leq s
\end{cases} \end{align}$$
By the theorem of Lehmann-Scheffé, the UMVUE for $\theta^m$ is, after operating the latter expression:
$$E[T|S=s]=\begin{cases}
0,  & if \ \ m>s \\
\frac{s!(n-m)!}{n!(s-m)!}, & if \ \ m\leq s
\end{cases}$$
