Let $a_0=5/2$ and $a_k=a^2_{k-1}-2$ for all $k\geq 1$.The question is to compute $$\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$
I tried to calculate few terms.$a_0=5/2$, $a_1=17/4,a_2=257/16$ it seems that $a_k$ is of the form $2^{2^k}+2^{-2^{k}}$ however I am not sure about it.How to proceed without doing much guesswork.Any ideas?
The explicit formula $a_k=2^{2^k}+2^{-2^{k}}$ is straightforward to prove by induction.
We also have
$$ a_k+1 = a_{k-1}^2-1 = (a_{k-1}-1)(a_{k-1}+1) $$
hence
$$ a_{k-1}-1 = \frac{a_k+1}{a_{k-1}+1} $$
and
$$ \prod_{k=1}^{N}\left(a_k-1\right)=\prod_{k=2}^{N+1}\left(a_{k-1}-1\right)=\frac{a_{N+1}+1}{a_1+1}. $$
Since
$$ \frac{x^{2^M}-y^{2^M}}{x-y} = \left(x^{2^{M-1}}+y^{2^{M-1}}\right)\cdots(x+y) $$
by picking $M=N+1, x=2$ and $y=\frac{1}{2}$ we get
$$a_N\cdot a_{N-1}\cdots a_0 =\frac{2^{2^{N+1}}-2^{-2^{N+1}}}{\frac{3}{2}}$$
and
$$ \prod_{k=0}^{N}\frac{a_k-1}{a_k}=\frac{3}{5}\cdot\frac{\prod_{k=1}^{N}(a_k-1)}{\prod_{k=1}^{N}a_k} = \frac{\left(\frac{3}{2}\right)^2}{a_1+1}\left(1+\frac{1}{2^{2^{N+1}}-2^{-2^{N+1}}}\right).$$
By letting $N\to \infty$ it follows that the wanted product equals $\frac{\left(\frac{3}{2}\right)^2}{a_1+1}= \color{red}{\large\frac{3}{7}}.$
