Answer
$xe^{x}=\Sigma_{n=0}^\infty\dfrac{x^{n+1}}{n!}$
Work Step by Step
Since, we know that the Maclaurin Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}$
Now, $xe^x=\Sigma_{n=0}^\infty \dfrac{(x)(x)^n}{n!}$
or, $xe^{x}=\Sigma_{n=0}^\infty\dfrac{x^{n+1}}{n!}$