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$\begin{aligned} f(x) & \le 2(1+x) & \forall x \in [1, \infty) \\ x f (x+1) & = (f(x))^2 - 1 & \forall x \in [ 1,\infty ) \end{aligned}$

A function $f:[1, \infty) \rightarrow [1, \infty)$ satisfy the conditions above. Find $f(2016)$.

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