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Given $\large \sqrt{\left \lceil 2 + \sqrt{\left \lceil 2+ \sqrt{ \left \lceil 2 + \sqrt{\left \lceil 2+ ... \right \rceil }\right \rceil }\right \rceil}\right \rceil } = \zeta$

find $\left \lfloor 1000 \zeta \right \rfloor$.

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