Let's solve the expression step-by-step.

Given: $\left(8^{-5} \cdot 4^{-4}\right)^{-1} \cdot \left(\frac{1}{16}\right)^6$

Step 1: Simplify the first part $(8^{-5} \cdot 4^{-4})^{-1}$

1. First, rewrite the expression inside the parentheses: $8^{-5} \cdot 4^{-4}$

2. Recall that $8 = 2^3$ and $4 = 2^2$, so rewrite the bases in terms of powers of 2: $8^{-5} = (2^3)^{-5} = 2^{-15}$ $4^{-4} = (2^2)^{-4} = 2^{-8}$

3. Multiply these together: $2^{-15} \cdot 2^{-8} = 2^{-23}$

4. Now, apply the $-1$ exponent: $(2^{-23})^{-1} = 2^{23}$

So, $\left(8^{-5} \cdot 4^{-4}\right)^{-1} = 2^{23}$.

Step 2: Simplify the second part $\left(\frac{1}{16}\right)^6$

1. Rewrite $16$ as a power of 2: $\frac{1}{16} = 2^{-4}$

2. Now, raise this to the sixth power: $\left(2^{-4}\right)^6 = 2^{-24}$

Step 3: Combine the results

Now we have: $2^{23} \cdot 2^{-24}$

Using the property of exponents $a^m \cdot a^n = a^{m+n}$: $2^{23 - 24} = 2^{-1} = \frac{1}{2}$

Final Answer

$\frac{1}{2}$

Would you like more details on any of these steps or have any questions?

Here are some related questions you might find helpful:

1. What happens when we multiply powers with the same base but different exponents?
2. How do negative exponents affect the base values in an expression?
3. Can we always convert numbers to powers of 2 or another base for simplification?
4. What are other methods to approach problems involving complex exponents?
5. How do fractional bases work with exponents?

Tip: Remember, rewriting numbers as powers of smaller primes (like 2, 3, or 5) often simplifies exponentiation problems.