To simplify the expression 

\[
\prod_{k=0}^{n+1} \sin\left(\frac{2\pi k}{n}\right)
\]

we can analyze it step by step.

### Step 1: Analyzing the structure
The given expression is a product of sine terms:

\[
\prod_{k=0}^{n+1} \sin\left(\frac{2\pi k}{n}\right)
\]

where \( k \) runs from 0 to \( n+1 \). Each sine term has the argument \( \frac{2\pi k}{n} \).

### Step 2: Special values of sine
For \( k = 0 \), we have:

\[
\sin\left(\frac{2\pi \cdot 0}{n}\right) = \sin(0) = 0
\]

Since one of the terms in the product is 0, the entire product evaluates to 0:

\[
\prod_{k=0}^{n+1} \sin\left(\frac{2\pi k}{n}\right) = 0
\]

### Final Simplified Expression:
\[
0
\]

Would you like to dive deeper into the reasons for this simplification or the role of other sine terms?

Here are some related questions:
1. What happens when we change the range of \( k \) in the product?
2. How does this product behave for different values of \( n \)?
3. Can we generalize this result for other trigonometric functions?
4. What if we exclude the \( k = 0 \) term from the product?
5. How does this product relate to Fourier series?

**Tip**: Identifying zeros in products or sums early can quickly simplify complex expressions!

Explore how to simplify the product of sine terms ∏_{k=0}^{n+1} sin(2πk/n), where one term evaluates to zero. This trigonometric expression simplifies to 0 using key concepts from trigonometry. Suitable for high school students in Grades 11-12.

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