HMMT November

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