Answer
$\Sigma_{n=0}^\infty (-1)^n(\dfrac{x^n}{n!})$
Work Step by Step
Consider the Maclaurin Series for $e^x$ as follows:
$e^x=\Sigma_{n=0}^\infty (\dfrac{x^n}{n!})$
Plug $-x$ instead of $x$.
This implies that $e^{-x}=\Sigma_{n=0}^\infty \dfrac{(-x)^n}{n!}=\Sigma_{n=0}^\infty (-1)^n(\dfrac{x^n}{n!})$