Expectation of a function in Poisson Distribution

Find the expectation of the function $\phi(x) = xe^{-x}$ in a Poisson distribution.

My Attempt: If $\lambda$ be the mean of Poisson distribution, then expectation of

$$\displaystyle \phi(x)=\sum_{x \mathop \ge 0} \frac{\phi(x)\lambda^xe^{-\lambda}}{x!}$$

$$= \displaystyle \sum_{x \mathop \ge 0} \frac{ xe^{-x}\lambda^xe^{-\lambda}}{x!}$$

$$= \displaystyle \lambda e^{-\lambda} \sum_{x \mathop \ge 1} \frac {e^{-x}} {\left({x-1}\right)!} \lambda^{x-1}$$

Now what?

Without the $e^{-x}$, the rest of summation is just a Taylor's expansion of $e^{\lambda}$, which gets cancelled.

But what do I do here?

Answer

Group together $e^{-x} \lambda^{x-1} = \frac{(\frac{\lambda}{e})^x}{\lambda}$, then do a bit of algebra to get the Taylor series for $e^s$ for some $s$.