1. **State the problem:** Find the domain of the function $$f(x) = \ln\left(\frac{1}{16^x} - 1\right)$$. 2. **Recall the domain rule for logarithms:** The argument of the natural logarithm must be strictly positive: $$\frac{1}{16^x} - 1 > 0$$ 3. **Solve the inequality:** $$\frac{1}{16^x} - 1 > 0 \implies \frac{1}{16^x} > 1$$ 4. Since $$16^x = (2^4)^x = 2^{4x}$$, rewrite the inequality: $$\frac{1}{2^{4x}} > 1 \implies 2^{-4x} > 1$$ 5. Because the base 2 is greater than 1, the inequality $$2^{-4x} > 1$$ holds when the exponent is greater than 0: $$-4x > 0 \implies x < 0$$ 6. **Conclusion:** The domain of $$f(x)$$ is all real numbers $$x$$ such that $$x < 0$$. **Final answer:** $$\boxed{(-\infty, 0)}$$