A Rather Simpl $e$ Sum

Recall the infinite series definition of $e^x$ : $e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots$ The sum seems very similar to the one being multiplied by $e^x$ . In fact, if we input $-x$ in place of $x$ : $e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^{n}}{n!} = \sum_{n=0}^{\infty} \frac{((x)(-1))^{n}}{n!} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}(-1)^{n}$ We get the sum in our expression! Substituting it with $e^{-x}$ we get $e^{x}e^{-x} = e^0 = \boxed{1}$