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
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 $x=1/e$.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$