Compute
\[\prod_{n = 0}^\infty \left[ 1 - \left( \frac{1}{2} \right)^{3^n} + \left( \frac{1}{4} \right)^{3^n} \right].\]
In general,
\[1 - x + x^2 = \frac{1 + x^3}{1 + x}.\]Thus,
\begin{align*}
\prod_{n = 0}^\infty \left[ 1 - \left( \frac{1}{2} \right)^{3^n} + \left( \frac{1}{4} \right)^{3^n} \right] &= \prod_{n = 0}^\infty \frac{1 + \left( \frac{1}{2} \right)^{3^{n + 1}}}{1 + \left( \frac{1}{2} \right)^{3^n}} \\
&= \frac{1 + \left( \frac{1}{2} \right)^3}{1 + \left( \frac{1}{2} \right)^0} \cdot \frac{1 + \left( \frac{1}{2} \right)^{3^2}}{1 + \left( \frac{1}{2} \right)^3} \cdot \frac{1 + \left( \frac{1}{2} \right)^{3^3}}{1 + \left( \frac{1}{2} \right)^{3^2}} \dotsm \\
&= \frac{1}{1 + \frac{1}{2}} = \boxed{\frac{2}{3}}.
\end{align*}