Answer
$1-5x+\dfrac{25x^2}{2}-\dfrac{125x^3}{6}+\dfrac{625x^4}{24}-...$
Work Step by Step
Consider the Maclaurin Series for $e^x$ as follows:
$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Plug $-5x$ instead of $x$ in the above series.
$e^{-5x}=1-5x+\dfrac{(-5x)^2}{2!}+\dfrac{(-5x)^3}{3!}-\dfrac{(-5x)^4}{4!}...$
$\implies 1-5x+\dfrac{25x^2}{2}-\dfrac{125x^3}{6}+\dfrac{625x^4}{24}-...$
