Answer
$e^{-x}=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^n}{n!}$
Work Step by Step
Since, we know that the Maclaurin Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}$
We need to replace in the above series $x$ with $-x$.
we have $e^{-x}=\Sigma_{n=0}^\infty \dfrac{(-x)^n}{n!}$
or, $e^{-x}=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^n}{n!}$
