, 2016

Solution:
Let $\{b_{i}\}_{i=0}^{\infty}$ be defined by $b_{n}=a_{n}-1$ and note that $b_{n}=b_{n-1}^{2}$. The infinite product is then
$$(1+b_{0})(1+b_{0}^{2})(1+b_{0}^{4}) \ldots (1+b_{0}^{2^{k}}) \ldots$$
By the polynomial identity
$$(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{k}}) \cdots = 1+x+x^{2}+x^{3}+\cdots = \frac{1}{1-x}$$
Our desired product is then simply
$$\frac{1}{1-(a_{0}-1)} = \frac{2}{3}$$