Answer
$e^{-e}=1-e+\frac{e^{2}}{2!}-\frac{e^{3}}{3!}+\frac{e^{4}}{4!}-....$
Work Step by Step
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 $e^{-e}=1-e+\frac{e^{2}}{2!}-\frac{e^{3}}{3!}+\frac{e^{4}}{4!}-....$
