1. **State the problem:** Simplify the expression $$\frac{\left(7^{-3}\right)^5^{-1}}{7^4 \cdot 7^{-5}} \cdot \frac{49^4}{49^{-5}}$$. 2. **Recall the rules:** - Power of a power: $$\left(a^m\right)^n = a^{m \cdot n}$$ - Product of powers with the same base: $$a^m \cdot a^n = a^{m+n}$$ - Quotient of powers with the same base: $$\frac{a^m}{a^n} = a^{m-n}$$ - Negative exponent: $$a^{-m} = \frac{1}{a^m}$$ - Express 49 as a power of 7: $$49 = 7^2$$ 3. **Simplify the numerator inside the first fraction:** $$\left(7^{-3}\right)^5 = 7^{-3 \cdot 5} = 7^{-15}$$ 4. **Apply the outer exponent -1:** $$\left(7^{-15}\right)^{-1} = 7^{-15 \cdot (-1)} = 7^{15}$$ 5. **Simplify the denominator of the first fraction:** $$7^4 \cdot 7^{-5} = 7^{4 + (-5)} = 7^{-1}$$ 6. **Simplify the first fraction:** $$\frac{7^{15}}{7^{-1}} = 7^{15 - (-1)} = 7^{16}$$ 7. **Rewrite the second fraction using base 7:** $$\frac{49^4}{49^{-5}} = \frac{(7^2)^4}{(7^2)^{-5}} = \frac{7^{8}}{7^{-10}} = 7^{8 - (-10)} = 7^{18}$$ 8. **Multiply the two results:** $$7^{16} \cdot 7^{18} = 7^{16 + 18} = 7^{34}$$ **Final answer:** $$7^{34}$$