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?

*

*You have shown that $T=\sum\limits_{i=1}^n X_i$ is a complete sufficient statistic for $\theta$.

*It is easy to verify that $E_{\theta}\left(\frac{T}{n}\right)=\theta$ and $E_{\theta}\left(\frac{T(T-1)}{n(n-1)}\right)=\theta^2$ for all $\theta\in(0,1)$.

*Inductively or otherwise one can show that 
$$E_{\theta}\left(\frac{T_{(k)}}{n_{(k)}}\right)=E_{\theta}\left(\frac{T(T-1)(T-2)\ldots(T-k+1)}{n(n-1)(n-2)\ldots(n-k+1)}\right)=\theta^k$$ for all $\theta\in(0,1)$ and for integer $k$ provided $0< k\le T\le n$.
Since $T\sim \mathsf{Bin}(n,\theta)$, we can also take some function $g$ of $T$ so that $g(T)$ is unbiased for $\theta^k$:
$$E_{\theta}\left[g(T)\right]=\sum_{j=0}^n g(j)\binom{n}{j}\theta^j(1-\theta)^{n-j} =\theta^k$$ 
Comparing coefficients we can solve for $g(\cdot)$. 
And the direct method outlined by @Akerbeltz is essentially the application of Rao-Blackwell theorem to find the UMVUE as $E\left[\prod\limits_{i=1}^k X_i\mid T\right]$.
Being a function of the complete sufficient statistic, $\frac{T_{(k)}}{n_{(k)}}$ is the UMVUE of $\theta^k$ by Lehmann-Scheffe theorem whenever $k$ is an integer with $0<k\le T\le n$. 
