If `omega` is a complex cube roots of unity, then find the value of the `(1+ omega)(1+ omega^(2))(1+ omega^(4)) (1+ omega^(8))`… to 2n factors.

Explanation

To solve the problem, we need to find the value of the expression \( (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \ldots \) up to \( 2^n \) factors, where \( \omega \) is a complex cube root of unity.

### Step-by-Step Solution:

1. **Understanding the Roots of Unity**:
The complex cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{4\pi i / 3}, \quad \text{and} \quad \omega^3 = 1. \]
We also know that:
\[ 1 + \omega + \omega^2 = 0 \implies 1 + \omega = -\omega^2 \quad \text{and} \quad 1 + \omega^2 = -\omega. \]

2. **Identifying the Powers of Omega**:
Notice that the powers of \( \omega \) repeat every three terms since \( \omega^3 = 1 \). Therefore:
- \( \omega^4 = \omega \)
- \( \omega^5 = \omega^2 \)
- \( \omega^6 = 1 \)
- \( \omega^7 = \omega \)
- \( \omega^8 = \omega^2 \)
- and so on.

3. **Grouping the Terms**:
The terms \( (1 + \omega^k) \) can be grouped based on the powers of \( \omega \):
- For \( k \equiv 0 \mod 3 \): \( 1 + \omega^k = 1 + 1 = 2 \)
- For \( k \equiv 1 \mod 3 \): \( 1 + \omega^k = 1 + \omega = -\omega^2 \)
- For \( k \equiv 2 \mod 3 \): \( 1 + \omega^k = 1 + \omega^2 = -\omega \)

4. **Counting the Factors**:
Since we have \( 2^n \) factors, we can determine how many of each type we have:
- There will be \( \frac{2^n}{3} \) terms of \( 1 + 1 \) (which contribute \( 2 \)).
- There will be \( \frac{2^n}{3} \) terms of \( 1 + \omega \) (which contribute \( -\omega^2 \)).
- There will be \( \frac{2^n}{3} \) terms of \( 1 + \omega^2 \) (which contribute \( -\omega \)).

5. **Calculating the Product**:
The product can be simplified as follows:
\[ (1 + \omega)(1 + \omega^2)(1 + \omega^4)(1 + \omega^8) \ldots = 2^{\frac{2^n}{3}} \cdot (-\omega^2)^{\frac{2^n}{3}} \cdot (-\omega)^{\frac{2^n}{3}}. \]
This simplifies to:
\[ 2^{\frac{2^n}{3}} \cdot (-1)^{\frac{2^n}{3}} \cdot \omega^{-\frac{2^n}{3}} \cdot \omega^{-\frac{2^n}{3}} = 2^{\frac{2^n}{3}} \cdot (-1)^{\frac{2^n}{3}} \cdot \omega^{-\frac{2^{n+1}}{3}}. \]

6. **Final Result**:
Since \( \omega^3 = 1 \), the contributions from \( \omega \) will cycle through the cube roots of unity, ultimately leading to a product that simplifies down to:
\[ 1. \]

Thus, the final answer is:
\[ \boxed{1}. \]