Consider a regular polygon with $2^n$ sides, for $n \ge 2$, inscribed in a circle of radius $1$. Denote the area of this polygon by $A_n$. Compute $\prod_{i=2}^{\infty}\frac{A_i}{A_{i+1}}$
1. **Understanding the Problem:**
   We are given a regular polygon with \(2^n\) sides inscribed in a circle of radius 1. We need to compute the infinite product \(\prod_{i=2}^{\infty}\frac{A_i}{A_{i+1}}\), where \(A_n\) denotes the area of the polygon with \(2^n\) sides.

2. **Area of a Regular Polygon:**
   The area \(A_n\) of a regular polygon with \(2^n\) sides inscribed in a circle of radius 1 can be computed using the formula:
   \[
   A_n = \frac{1}{2} \cdot 2^n \cdot \sin\left(\frac{2\pi}{2^n}\right)
   \]
   Simplifying, we get:
   \[
   A_n = 2^{n-1} \sin\left(\frac{2\pi}{2^n}\right)
   \]

3. **Telescoping Product:**
   We need to compute the product \(\prod_{i=2}^{\infty}\frac{A_i}{A_{i+1}}\). Using the formula for \(A_n\), we have:
   \[
   \frac{A_i}{A_{i+1}} = \frac{2^{i-1} \sin\left(\frac{2\pi}{2^i}\right)}{2^i \sin\left(\frac{2\pi}{2^{i+1}}\right)} = \frac{1}{2} \cdot \frac{\sin\left(\frac{2\pi}{2^i}\right)}{\sin\left(\frac{2\pi}{2^{i+1}}\right)}
   \]

4. **Simplifying the Product:**
   The product then becomes:
   \[
   \prod_{i=2}^{\infty} \frac{A_i}{A_{i+1}} = \prod_{i=2}^{\infty} \left( \frac{1}{2} \cdot \frac{\sin\left(\frac{2\pi}{2^i}\right)}{\sin\left(\frac{2\pi}{2^{i+1}}\right)} \right)
   \]
   Notice that this product telescopes. Specifically, the \(\sin\left(\frac{2\pi}{2^{i+1}}\right)\) term in the denominator of one fraction will cancel with the \(\sin\left(\frac{2\pi}{2^i}\right)\) term in the numerator of the next fraction.

5. **Telescoping Effect:**
   After cancellation, we are left with:
   \[
   \prod_{i=2}^{\infty} \frac{A_i}{A_{i+1}} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdots = \frac{1}{2^{\infty}} = 0
   \]
   However, this is incorrect because we need to consider the initial terms more carefully.

6. **Correct Telescoping:**
   The correct telescoping product should be:
   \[
   \prod_{i=2}^{\infty} \frac{A_i}{A_{i+1}} = \frac{A_2}{A_{\infty}}
   \]
   where \(A_{\infty}\) is the area of the circle, which is \(\pi\).

7. **Initial Area Calculation:**
   For \(n=2\), the polygon is a square, so:
   \[
   A_2 = 2^{2-1} \sin\left(\frac{2\pi}{2^2}\right) = 2 \sin\left(\frac{\pi}{2}\right) = 2
   \]

8. **Final Product:**
   Therefore, the product is:
   \[
   \prod_{i=2}^{\infty} \frac{A_i}{A_{i+1}} = \frac{A_2}{A_{\infty}} = \frac{2}{\pi}
   \]

The final answer is \(\boxed{\frac{2}{\pi}}\)