Let $\{X_i\}_{i=1}^{n}$ be iid according to the Poisson distribution P(λ). Use Method 1 to find the UMVU estimator of $\lambda^k$ for any positive integer k.
Method 1: If T is a complete sufficient statistic, the UMVU estimator of any U-estimable function $g(θ)$ is uniquely determined by the set of equations $E_{θ}[\delta(T)] = g(θ)$ for all $\theta$:
This is my attempt:
$\sum_{t=0}^{\infty}\delta(t)\frac{\lambda^t e^{-\lambda}}{t!}=\lambda^k$
through algebra equivalent to
$\sum_{t=0}^{\infty}\delta(t)\frac{\lambda^t}{t!}=\sum_{n=0}^{\infty}\frac{\lambda^{k+n}}{n!}$
After this I get stuck trying to manipulate the two series to compare coefficients to identify $\delta(t)$. Any help would be appreciated.
$T=\sum\limits_{i=1}^{n} X_i$ is a complete sufficient statistic in this case. $T$ is a Poisson random variable with mean $n \lambda$. Let us find a nice function $g$ such that $$\mathbb{E}[g(T)]=\lambda^{k}$$. $\sum\limits_{t=0}^{\infty} g(t)e^{-n \lambda} \frac{n^t\lambda^{t}}{t!}=\lambda^{k}$. so $$\sum\limits_{t=0}^{\infty} g(t) \frac{n^{t}\lambda^t}{t!}=\lambda^{k} e^{n \lambda}$$, recall that $$\lambda^{k}e^{n \lambda}=\sum\limits_{t=0}^{\infty} \frac{n^{t} \lambda^{k+t}}{t!}=\sum\limits_{u=k}^{\infty} n^{u-k} \frac{\lambda^{u}}{(u-k)!}. $$Hence $g(t)=0$ for $0 \le t \le k-1$. What can you say about $g(t)$ for $t \ge k$?