Answer
$x+x^2+\dfrac{1}{2!}x^3+\dfrac{1}{3!}x^4+\dfrac{1}{4!}x^5+...$
Work Step by Step
Consider the Maclaurin Series for $e^x$ is defined as:
$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Multiply the above series with $x$ as follows:
Then $xe^x=x(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...)$
$\implies x+x^2+\dfrac{1}{2!}x^3+\dfrac{1}{3!}x^4+\dfrac{1}{4!}x^5+...$
