Using the identity $$(x + x^{-1})^2 - 2 = x^2 + x^{-2},$$ we may check by induction on $k$ that $a_k = 2^{2^k} + 2^{-2^k}$; in particular, the product is absolutely convergent. Using the identities $$\frac{x^2 + 1 + x^{-2}}{x + 1 + x^{-1}} = x - 1 + x^{-1},$$ $$\frac{x^2 - x^{-2}}{x - x^{-1}} = x + x^{-1},$$ we may telescope the product to obtain $$\prod_{k=0}^\infty \left( 1 - \frac{1}{a_k} \right) = \prod_{k=0}^\infty \frac{2^{2^k} - 1 + 2^{-2^k}}{2^{2^k} + 2^{-2^k}} = \prod_{k=0}^\infty \frac{2^{2^{k+1}} + 1 + 2^{-2^{k+1}}}{2^{2^k} + 1 + 2^{-2^k}} \cdot \frac{2^{2^k} - 2^{-2^k}}{2^{2^{k+1}} - 2^{2^{-k-1}}} = \frac{2^{2^0} - 2^{-2^0}}{2^{2^0}+1 + 2^{-2^0}} = \frac{3}{7}.$$