Why does the minimum value of $x^x$ equal $1/e$?

Hint: Note that $x^x = e^{x\log x}$.

So minimizing $x^x$ is the same as minimizing $x\log x$

Let $f(x)=x^x$. Note that $f(x)$ is only defined for $x>0$.

Then $$ \ln f(x)=x\cdot\ln x\tag{1} $$ Differentiating (1) gives $$ \frac{1}{f(x)}f^\prime(x)=x\frac{1}{x}+\ln x=1+\ln x $$ Note that we have used the chain rule and the product rule.

Solving for $f^\prime(x)$ gives $$ f^\prime(x)=f(x)(1+\ln x)=x^x(1+\ln x) $$ Can you use this to locate the critical points of $f(x)$?

Hint: actually you are looking for a local/global minimum .. so look at the derivative of the function $f(x) = x^x$ $$f'(x) = x^x (\log (x)+1)$$ which equals $0 \iff x = \frac{1}{e}$