Recurrence relation of type $a_{n+1} = a^2_{n}-2a_{n}+2$ 
A sequence $\{a_{n}\}$ is defined by $a_{n+1} = a^2_{n}-2a_{n}+2\forall n\geq 0$ and $a_{0} =4$
And another sequence $\{b_{n}\}$ defined by the formula $\displaystyle b_{n} = \frac{2b_{0}b_{1}b_{2}..........b_{n-1}}{a_{n}}\forall n\geq 1$ and $\displaystyle  b_{0}=\frac{1}{2}$,Then 
$(a)$ value of $a_{10}$
$(b)\;\; $ The value of $n$ for which $\displaystyle b_{n} = \frac{3280}{3281}$
$(c)$ The Sequence $\{b_{n}\}$ satisfy the recurrence formula
$\bf{Options::}$
$(1)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1-b^2_{n}}\;\;\;\;\;\; (b)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1b^2_{n}}\;\;\;\;\;\; (c)\; \displaystyle b_{n+1} = \frac{b_{n}}{1+2b^2_{n}}\; (d)\; \displaystyle b_{n+1} = \frac{b_{n}}{1-2b^2_{n}}$

$\bf{My\; Try::}$ Given $a_{n+1} = a^2_{n}-2a_{n}+2 = \left(a_{n}-1\right)^2+1$
So $a_{1} = (a_{0}-1)^2+1=(4-1)^2+1=10$
Similarly $a_{2} = (a_{1}-1)^2+1 = 9^2+1 = 82$
Similarly $a_{3} = (81)^2+1 = $
But Calculation like this is very complex, Plz help me how can i solve above problems, Thanks
 A: Define $c_{n}=a_{n}-1$. Note that $c_{n+1}=c_{n}^{2}$. Then,
$c_{4} = c_{3}^{2} = c_{2}^{4} = c_{1}^{8} = c_{0}^{16} = 3^{16}$, so $a_{4}=3^{16}+1$.
A: For the first problem, as LaloVelasco answered, you have $$c_{n+1}=c_{n}^{2}$$ Take logarithms $$\log(c_{n+1})=2\log(c_n)$$ Define $d_n=\log(c_n)$ which makes $$d_{n+1}=2d_n\implies d_n=k\, 2^{n-1}\implies c_n=e^{k\, 2^{n-1}}\implies a_n=1+e^{k\, 2^{n-1}}$$ and $k=2 \log (a_0-1)$ which makes $$a_n=(a_0-1)^{2^n}+1$$ and then the result for any $n$.
Then, $a_{10}=(a_0-1)^{1024}+1$.
A: For the second recurrence
$$
b_{\,n + 1}  = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {b_{\,k} } }}
{{b_{\,n} }}
$$
we have
$$
\left\{ \begin{gathered}
  b_{\,0}  = 1/2 \hfill \\
  b_{\,1}  = \frac{2}
{{b_{\,0} }} = 4 \hfill \\
  b_{\,n + 1}  = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {b_{\,k} } }}
{{b_{\,n} }} = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {b_{\,k} } }}
{{\frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 2} {b_{\,k} } }}
{{b_{\,n - 1} }}}} = b_{\,n - 1} ^2  \hfill \\ 
\end{gathered}  \right.
$$
assuming that, in deducing $b_1$, we can adopt empty product = $1$
