Finding UMVUE of a function of parameter belonging to Poisson distribution Let $X_1, ..., X_n$ be iid from the Poisson ($\theta$) distribution. 
I have proven that $T = \sum_{i=1}^{n} x_i$ is the complete and sufficient statistic and it has a Poisson($n\theta$) distribution. I need to find the UMVUE of $\theta^k$, k>0. 
Attempts: 
I have noticed that $E(x_1x_2...x_k) = \theta^k$ due to the iid property and they are the unbiased estimator. 
I tried using the Lehmann Scheffe Theorem 
$\begin{equation} 
\sum_{t=0}^{n} g(t)\frac{e^{-n\theta}{(n\theta)}^t}{t!} = \theta^k\end{equation}$ but arrived with the result of $g(t) = \theta^k$, which does not make sense. I tried using factorial moments, but think it is too complicated when k is not specified. 
Can anyone point me in the right direction? 
Thank you. 
 A: As you mentioned, $$P(T=t)=\frac{e^{-n\theta} (n\theta)^t}{t!} $$
Let $I(T)$ be a function such that 
\begin{align}
I(T)=\begin{cases}\displaystyle\frac{T!}{(T-k)!}&,\text{ when }T\geq k\\0 &,\text{ elsewhere }\end{cases} 
\end{align}
Then,
$$E\left[  I(T) \right]=\sum_{t=k}^{\infty} \frac{e^{-n\theta} (n\theta)^{t-k}}{(t-k)!} n^k \theta^k =n^k\theta^k$$
So, $$E\left[ \frac{I(T)}{n^k} \right]=\theta^k$$
And we know that $T$ is the complete and sufficient statistic, so the function of $T$ given by $I(T)/n^k$ is the UMVUE of $\theta^k$.
A: Try comparing power series coefficients (warning: this can lead to very complicated calculations).
Specifically, assuming that there is a UMVUE $\delta$, we know it is unbiased and a function of $T$, so we can write that $\delta(X) = \eta(T(X))$, and that $\mathbb{E}_{\theta}[\eta(T(X))] = \theta^k$, i.e. that $$\sum_{t=0}^{\infty} \frac{\theta^t n^t e^{-n\theta}}{t!}\eta(t) = \sum_{t=0}^{\infty}\mathbb{P}_{\theta}(T = t) \eta(t) = \theta^k \,.$$
Alternatively, (by Theorem 4.4. in Keener, Theoretical Statistics: Topics for a Core Course - proof involves Rao-Blackwell - is this called Lehmann-Scheffe in the literature?) we also can say that: $$\eta(T(X)) = \mathbb{E}_{\theta}[\delta(X)|T] $$ This method is even more general, but for me at least is even more difficult to use in practice.
