## Calculus 8th Edition

$e^{-e}=1-e+\frac{e^{2}}{2!}-\frac{e^{3}}{3!}+\frac{e^{4}}{4!}-....$
Given: $1-e+\frac{e^{2}}{2!}-\frac{e^{3}}{3!}+\frac{e^{4}}{4!}-....$ Remember that $e^{x}= 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+....$ Re-write the given series as $1+(-e)+\frac{(-e)^{2}}{2!}+\frac{(-e)^{3}}{3!}+\frac{(-e)^{4}}{4!}+....$ Therefore, this is equal to $e^{-e}=1-e+\frac{e^{2}}{2!}-\frac{e^{3}}{3!}+\frac{e^{4}}{4!}-....$