Answer
$\Sigma_{n=0}^\infty\dfrac{x^{n+1}}{n!}$
Work Step by Step
Consider the Maclaurin Series for $e^x$ as follows:
$e^x=\Sigma_{n=0}^\infty (\dfrac{x^n}{n!})$
Multiply the above series with $x$
we get $xe^x=\Sigma_{n=0}^\infty \dfrac{x \cdot x^n}{n!}=\Sigma_{n=0}^\infty\dfrac{x^{n+1}}{n!}$
