What is the value of $e^\frac{-1}{e}$? As the title suggests, what is the value of $e^\frac{-1}{e}$?? I never understood how things to the power of irrational numbers are calculated. The more broad question would be what is the minimum value for the function $y=x^x$, such that $x \geq 0$? I have tried taking the derivative to get $x^x(\ln(x)+1) = 0$, then simplifying to get $\ln(x)+1=0$, and $\ln(x)=-1$, so $x = \frac{1}{e}$. This simplifies to $e^\frac{-1}{e}$. What do I put for my final answer? Any help would be appreciated.
 A: As the replies indicate, you should put $e^{\frac{-1}{e}}$ since there is no simpler form.
Nevertheless, you seem to seek an intuition into irrational exponents, namely $\frac{-1}{e}$.
The basic idea is that every real number is the limit of a sequence of rational numbers, in this case with the help of the Taylor series of $e^x$
$$\frac{-1}{e}=-e^{-1}=-\sum_{n=0}^\infty \frac{(-1)^n}{n!}
= \sum_{n=0}^\infty \frac{(-1)^{n+1}}{n!}
=-1+\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\cdots$$
Which follows
$$ \exp\left(\frac{-1}{e}\right)
= \exp\left(\sum\limits_{n=0}^\infty \frac{(-1)^{n+1}}{n!} \right)
= \prod\limits_{n=0}^\infty \exp\left( \frac{(-1)^{n+1}}{n!} \right)
= e^{-1}\cdot e^{\frac{1}{1!}}\cdot e^{-\frac{1}{2!}}\cdot e^{\frac{1}{3!}}\cdot e^{-\frac{1}{4!}}\cdot\dots
$$
Which translates into a product of roots of $e$ and $\frac{1}{e}$.
$$e^{\frac{-1}{e}} = e^{-1}\cdot e\cdot \sqrt{e^{-1}}\cdot \sqrt[3!]{e}\cdot
\sqrt[4!]{e^{-1}}\cdot \ldots$$
While the obtained expression might make sense, it is not mathematically significant.
However, one might use this expression to write a small program to compute $e^\frac{-1}{e}$ in the desired precision.
Check out Real exponents on Wikipedia.
