Get it's minimum

Let $y(x) = x^x$ .

$\begin{aligned} y & = x^x = e^{x\ln x} \\ \frac {dy}{dx} & = (\ln x + 1)e^{x\ln x} = (\ln x + 1)x^x\end{aligned}$ .

Equating $\dfrac {dy}{dx} = 0$ , since $x^x > 0$ implies $\ln x + 1 = 0$ or $\ln x = -1$ $\implies x = e^{-1} = \dfrac 1e$ .

Now, note that $\dfrac {d^2 y}{dx^2} = \dfrac {x^2}x + (\ln x+1)^2 x^x > 0$ , when $x=\dfrac 1e$ . This means that $y \left(\frac 1e\right) =\boxed{e^{-\frac 1e}}$ is minimum.