The minimum of x^x

Limit. $x\ln x\to0$ as $x\to0^+$, so $f(x)\to e^0=1$: $f$ extends continuously to $0$ with $f(0)=1$.

Derivative. $(x\ln x)'=\ln x+1$, so $f'(x)=x^x(\ln x+1)$. It is negative for $x<\frac1e$ and positive for $x>\frac1e$.

Variations. $f$ decreases from $1$ (at $0^+$) to its minimum, then increases to $+\infty$:
$$f_{\min}=f\left(\frac1e\right)=e^{-1/e}\approx0.692.$$
Also $f(1)=1$, so $f$ takes the value $1$ again at $x=1$.

Consequence. For every $x>0$, $x^x\ge e^{-1/e}$, with equality only at $x=\frac1e$. Also $f(x)\to+\infty$ as $x\to+\infty$ (faster than any exponential $a^x$, since $\frac{x^x}{a^x}=\left(\frac xa\right)^x\to\infty$).

Remark. Since $f'(x)\to-\infty$ as $x\to0^+$, the curve leaves the point $(0,1)$ with a vertical tangent, heading downward.