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!