Answer
$x+x^2+\dfrac{x^3}{2!}+\dfrac{x^4}{3!}+\dfrac{x^5}{4!}+...$
Work Step by Step
Since, we know that the Maclaurin Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Now, $xe^x=x(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...)$
or, $=x+x^2+\dfrac{x^3}{2!}+\dfrac{x^4}{3!}+\dfrac{x^5}{4!}+...$
