Calling on all Math Whizzes... Can anyone please help...

Show that for $\displaystyle \lambda>0$

$\displaystyle \sum_{i=1}^{\infty }i \times \frac{e^{-\lambda}\lambda^{i}}{i!} = \lambda$

where

$\displaystyle x!=\prod_{j=0}^{x=1}(x-j)=x \times (x-1)\times ... \times 1, x=1,2,3....$

$\displaystyle and$

$\displaystyle x! = 1 , x=0$