1. **State the problem:** Simplify the expression $\left(\frac{1}{4}\right)^{-3}$.\n\n2. **Recall the rule for negative exponents:** For any nonzero number $a$ and integer $n$, $a^{-n} = \frac{1}{a^n}$.\n\n3. **Apply the negative exponent rule:** $$\left(\frac{1}{4}\right)^{-3} = \frac{1}{\left(\frac{1}{4}\right)^3}$$\n\n4. **Calculate the cube of $\frac{1}{4}$:** $$\left(\frac{1}{4}\right)^3 = \frac{1^3}{4^3} = \frac{1}{64}$$\n\n5. **Substitute back:** $$\left(\frac{1}{4}\right)^{-3} = \frac{1}{\frac{1}{64}}$$\n\n6. **Simplify the complex fraction:** $$\frac{1}{\frac{1}{64}} = 1 \times \frac{64}{1} = 64$$\n\n**Final answer:** $64$