# Expectation of an estimator following a Poisson distribution

I have the parameter $\theta=exp(-\lambda)$. I want to show that the estimator

$R=(1-\frac{1}{n})^T$ is an unbiased estimator of $\theta$, where $T=\frac{1}{n}\sum_{i=1}^{n}X_i$ and it follows a Poisson distribution with expectation $n\lambda$

I know that $E(R)=exp(-\lambda)$ but I can't seem to get to this answer.

I have this:

$E(R)=E((1-\frac{1}{n})^T)$

But I'm not even sure where to go next!

Hint 1: By definition of the expectation, $$\mathbb{E}_\lambda\left[\left(1-\frac{1}{n}\right)^T\right]=\sum_{m=0}^\infty\left(1-\frac{1}{n}\right)^m \mathbb{P}_\lambda(T=m)=\sum_{m=0}^\infty\left(1-\frac{1}{n}\right)^m\, \frac{(n\lambda)^m}{m!}\,\exp\{-n\lambda\}$$ using the law of the unconscious statistician
Hint 2: By definition of the moment generating function $$\Psi_\lambda(t)=\mathbb{E}_\lambda\left[\exp\{tT\}\right]$$ and $$\mathbb{E}_\lambda\left[\left(1-\frac{1}{n}\right)^T\right]=\Psi_\lambda\left\{\log\left(1-\frac{1}{n}\right)\right\}$$