Answer
$ \Sigma_{k=0}^{\infty} \dfrac{x^{k+1}}{k!}$
Work Step by Step
$e^x=\Sigma_{k=0}^{\infty} \dfrac{x^k}{k!}$
We will multiply with $x$.
$x e^x=\Sigma_{k=0}^{\infty} x \cdot \dfrac{x^k}{k!}$
So, $x \ e^{x}= \Sigma_{k=0}^{\infty} \dfrac{x^{k+1}}{k!}$
