1. **State the problem:** Calculate the limit $$\lim_{x \to 0^+} \frac{x^x - 1}{\ln(x + 1)}.$$\n\n2. **Recall important formulas and rules:**\n- The expression $x^x$ can be rewritten as $e^{x \ln x}$.\n- For small $x$, $\ln(1 + x) \approx x$.\n- Use expansions for limits involving indeterminate forms like $0/0$.\n\n3. **Rewrite the numerator:**\n$$x^x = e^{x \ln x}.$$\nAs $x \to 0^+$, $x \ln x \to 0$, so use the expansion:\n$$e^{x \ln x} = 1 + x \ln x + \frac{(x \ln x)^2}{2} + \cdots.$$\nThus,\n$$x^x - 1 \approx x \ln x.$$\n\n4. **Rewrite the denominator:**\nFor small $x$,\n$$\ln(x + 1) \approx x.$$\n\n5. **Form the limit expression:**\n$$\lim_{x \to 0^+} \frac{x^x - 1}{\ln(x + 1)} \approx \lim_{x \to 0^+} \frac{x \ln x}{x} = \lim_{x \to 0^+} \ln x.$$\n\n6. **Evaluate the limit:**\nAs $x \to 0^+$, $\ln x \to -\infty$.\n\n**Final answer:**\n$$\lim_{x \to 0^+} \frac{x^x - 1}{\ln(x + 1)} = -\infty.$$