Chapter 9.10, Problem 1QCE

The Maclaurin series for $e^{-x}{}^{{}^2}$ obtained by substituting $-x^2$ for $x$ in the series $e^x={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{x^k}{k!}}\text{is}{e}^{-x^2}={\displaystyle {\sum }_{k=0}^{\infty }\text{\_\_\_\_\_\_\_}\text{.}}$