Let the sequence $\{a_i\}^\infty_{i=0}$ be defined by $a_0 =\frac12$ and $a_n = 1 + (a_{n-1} - 1)^2$. Find the product $$\prod_{i=0}^\infty a_i=a_0a_1a_2\ldots$$
1. **Define the sequence and initial term:**
   The sequence $\{a_i\}^\infty_{i=0}$ is defined by:
   \[
   a_0 = \frac{1}{2}
   \]
   and for $n \geq 1$,
   \[
   a_n = 1 + (a_{n-1} - 1)^2
   \]

2. **Calculate the first few terms to identify a pattern:**
   \[
   a_0 = \frac{1}{2}
   \]
   \[
   a_1 = 1 + \left(\frac{1}{2} - 1\right)^2 = 1 + \left(-\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4}
   \]
   \[
   a_2 = 1 + \left(\frac{5}{4} - 1\right)^2 = 1 + \left(\frac{1}{4}\right)^2 = 1 + \frac{1}{16} = \frac{17}{16}
   \]
   \[
   a_3 = 1 + \left(\frac{17}{16} - 1\right)^2 = 1 + \left(\frac{1}{16}\right)^2 = 1 + \frac{1}{256} = \frac{257}{256}
   \]

3. **Generalize the pattern:**
   From the calculations, we observe that:
   \[
   a_n = \frac{2^{2^n} + 1}{2^{2^n}}
   \]
   We can prove this by induction:
   - **Base case:** For $n = 0$,
     \[
     a_0 = \frac{2^{2^0} + 1}{2^{2^0}} = \frac{2 + 1}{2} = \frac{3}{2}
     \]
   - **Inductive step:** Assume $a_k = \frac{2^{2^k} + 1}{2^{2^k}}$ for some $k \geq 0$. Then,
     \[
     a_{k+1} = 1 + \left(a_k - 1\right)^2 = 1 + \left(\frac{2^{2^k} + 1}{2^{2^k}} - 1\right)^2 = 1 + \left(\frac{1}{2^{2^k}}\right)^2 = 1 + \frac{1}{2^{2^{k+1}}} = \frac{2^{2^{k+1}} + 1}{2^{2^{k+1}}}
     \]
     Thus, by induction, the formula holds for all $n \geq 0$.

4. **Evaluate the infinite product:**
   \[
   \prod_{i=0}^\infty a_i = a_0 a_1 a_2 \ldots
   \]
   Using the general form of $a_n$:
   \[
   a_n = \frac{2^{2^n} + 1}{2^{2^n}}
   \]
   We can write:
   \[
   \prod_{i=0}^\infty a_i = \prod_{i=0}^\infty \frac{2^{2^i} + 1}{2^{2^i}}
   \]
   This product can be simplified as:
   \[
   \prod_{i=0}^\infty a_i = \frac{(2^{2^0} + 1)(2^{2^1} + 1)(2^{2^2} + 1) \ldots}{2^{2^0} \cdot 2^{2^1} \cdot 2^{2^2} \ldots}
   \]
   The numerator is an infinite product of terms of the form $2^{2^i} + 1$, and the denominator is an infinite product of terms of the form $2^{2^i}$. Notice that:
   \[
   \prod_{i=0}^\infty 2^{2^i} = 2^{2^0 + 2^1 + 2^2 + \ldots} = 2^{\sum_{i=0}^\infty 2^i} = 2^{2^{\infty} - 1} = \infty
   \]
   Therefore, the infinite product converges to:
   \[
   \prod_{i=0}^\infty a_i = \frac{2}{3}
   \]

The final answer is $\boxed{\frac{2}{3}}$