Show that $(1-\frac{1}{n})^{\sum_{i=1}^n X_i}$ is an unbiased estimator of $\tau(\theta)=e^{-\lambda}$

Question

Let $S=\sum_{i=1}^n X_i$ than show that $T(X_1,\ldots,X_n)= \left(1-\frac{1}{n}\right)^{\sum_{i=1}^n X_i}$ is an unbiased estimator of $\tau(\theta)=e^{-\lambda}$ where $X_1,\ldots,X_n$ are IID from POI($\lambda$) with $n>1$.

Answer 1

Hint: Try using the fact that $S= \sum X_i$ is a Poisson random variable with parameter $n\lambda$ so that $$E\left[\left(1-\frac{1}{n}\right)^S\right]=\sum_{i=0}^\infty \left(1-\frac{1}{n}\right)^i e^{-n\lambda}\frac{(n\lambda)^i}{i!}=e^{-n\lambda}\sum_{i=0}^\infty \frac{(n\lambda - \lambda)^i}{i!}$$

Answer 2

So basically, you can say that since you're looking for an unbiased estimator, you need $$E[\hat{\tau(\theta)}]=\tau(\theta)=e^{-\lambda}$$ So we know $$E[\hat{\tau(\theta)}]=E\left[\left(1-\frac{1}{n}\right)^S\right]$$ And the sum of n random IID Poisson variables is just another Poisson r.v with parameter $n\lambda$. Then using the definition of expectation you get your answer.