Let the sequence ${\left\{a_i\right\}}_{i=0}^{\infty }$ be defined by $a_0=\frac{1}{3}$ and  $a_n=1+{(\left(a_{n-1}\right)-1)}^2$. The value of the product  $\prod _{i=0}^{\infty }{a}_i (where\prod _{i=0}^{\infty }{a}_i=a_0{a}_1{a}_2\mathrm{......})$  is ______

Let ${\left\{b_i\right\}}_{i=0}^{\infty }$ be defined by $b_n=a_n-1$ and note $b_n=b_{n-1}^2$. The infinite product is then  $(1+b_0)(1+b_0^2)(1+b_0^4)\mathrm{........}(1+b_0^{2^k})\mathrm{..........}$
By the polynomial identity
 $(1+x)(1+x^2)(1+x^4)\mathrm{.......}(1+x^{2^k})\mathrm{....}=1+x+x^2+x^3+\mathrm{.......}=\frac{1}{1-x}$
Our desired product is then simply
$\frac{1}{1-(a_0-1)}=\frac{1}{2-a_0}=\frac{3}{5}$