So the title says almost everything.
Why is the minimum of the function $x\mapsto x^x$ exactly at $x= e^{-1}$?
Don't get me wrong. I am able to write $x^x$ as $\exp(x \ln(x))$ myself and can differentiate $x^x$ to conclude that the minimum lies at $x=e^{-1}$.
However, since $x^x$ can be defined without the $e$-function I guess there has to be another explaination which hopefully gives more insight.
An alternative would be to study the extrema of the function $x^y$ subjected to the condition $x-y=0$ using the method of Langrange multipliers.