How do I get from $\sum_{n=0}^{\infty}\frac{(1/e)^n}{n!} = e^{1/e}$

How do I get from $$\sum_{n=0}^{\infty}\frac{(1/e)^n}{n!} = e^{1/e}$$

I am given

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$

I am thinking

$$\sum \frac{n!}{n^n}\cdot \frac{1}{n!}$$

But it seems wrong

-
Let $x=1/e$.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
But then how do I get from $\sum{\frac{x^n}{n!}}=x$ then? –  Jiew Meng Apr 20 '12 at 10:23
@JiewMeng: Huh? That is simply not true. Look at the equation $\sum \frac{x^n}{n!} = e^x$ again. –  Martin Wanvik Apr 20 '12 at 10:28
@Jiew - loosely speaking, the formula Martin pointed to says that - $\sum \frac{something^n}{n!} = e^{something}$ –  Amihai Zivan May 22 '12 at 6:45