How to show $\sum_0^\infty \frac{x\lambda^x} {x!} = \lambda e^\lambda$? I know that $\sum_0^\infty \frac{\lambda^x} {x!} = e^\lambda$, but I'm having a really difficult time dealing with the extra $x$.

Very inelegantly and in schoolboy fashion we can write
\begin{align}
\sum_{x=0}^\infty x\frac{\lambda^x}{x!}
&= 0\frac{\lambda^0}{0!} + 1\frac{\lambda^1}{1!} + 2\frac{\lambda^2}{2!}
+ 3\frac{\lambda^3}{3!} + 4\frac{\lambda^4}{4!} + \cdots\\
&= 0 + \lambda +\frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!}+\cdots\\
&= \lambda +\frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!}+\cdots\\
&= \lambda\left[1 + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!}
+ \frac{\lambda^3}{3!} + \cdots\right]\\
&= \lambda e^\lambda
\end{align}
