To solve problem 1, we need to simplify the expression given:

$\frac{(2^{2n})^2 \cdot (4^n)^{n-1}}{(2^n)^{2n+1}}$

Let's break it down step by step.

Step 1: Simplify the Numerator

The numerator is $(2^{2n})^2 \cdot (4^n)^{n-1}$.

1. $(2^{2n})^2$: $(2^{2n})^2 = 2^{2n \cdot 2} = 2^{4n}$

2. $(4^n)^{n-1}$: $4^n = (2^2)^n = 2^{2n}$ $(2^{2n})^{n-1} = 2^{2n \cdot (n-1)} = 2^{2n^2 - 2n}$

So the numerator becomes: $2^{4n} \cdot 2^{2n^2 - 2n} = 2^{4n + 2n^2 - 2n} = 2^{2n^2 + 2n}$

Step 2: Simplify the Denominator

The denominator is $(2^n)^{2n+1}$.

1. $(2^n)^{2n+1}$: $(2^n)^{2n+1} = 2^{n \cdot (2n+1)} = 2^{2n^2 + n}$

Step 3: Combine Numerator and Denominator

Now we have: $\frac{2^{2n^2 + 2n}}{2^{2n^2 + n}}$

Using the exponent subtraction rule $\frac{a^m}{a^n} = a^{m-n}$, we get: $2^{(2n^2 + 2n) - (2n^2 + n)} = 2^{2n^2 + 2n - 2n^2 - n} = 2^n$

So, the simplified form of the given expression is: $2^n$

Do you want details or have any questions about this solution?

Here are some questions you might want to ask:

1. How do I simplify expressions with exponents?
2. What are the properties of exponents used in simplification?
3. How do you handle negative exponents in simplification?
4. Can you explain how to convert bases during simplification?
5. What strategies can be used to verify the simplification of complex expressions?

Tip: When simplifying expressions with exponents, always look for opportunities to apply the rules of exponents, such as $a^m \cdot a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$.