Number theory

Let $P = 1!\cdot 2! \cdot 3! \cdots 2015! \cdot 2016! \\ P = 1^{2016}\cdot 2^{2015}\cdot 3^{2014} \cdots 2015^{2} \cdot 2016^1 \\ P= 2\cdot 4\cdot 6\cdots 2014\cdot 2016 \left( 1^{2016}\cdot 2^{2014}\cdot 3^{2014} \cdots 2015^{2} \right) \\ P = 1008!\cdot \underbrace{2^{1008}\left( 1^{2016}\cdot 2^{2014}\cdot 3^{2014} \cdots 2015^{2} \right)}_{\text{Perfect Square}}$

Our Answer : $\boxed{1008!}$

The only possible solution to this problem is when the last number in the sequence is a multiple of 4. And the solution is removing factorial of n/2 to get a perfect square. If n/2 is also a perfect square, factorial of (n/2 - 1) is also a solution.

Example:

1!.2!.3!.4!.5!.6!.7!.8! = (2^23).(3^9).(5^4).(7^2) is not a perfect square, but when we remove 4! the resulting product = (2^20).(3^8).(5^4).(7^2) becomes a perfect square. Since 4 is also a perfect square, if we remove 3! we also get a perfect square which is (2^22).(3^8).(5^4).(7^2).

In the case 1!.2!.3!.4!.5!.6!.7!.8!.9!.10!.11!.12!.13! = (2^66).(3^31).(5^13).(7^7).(11^3).(13^1) there is no single number whose factorial we can remove that will make it a perfect square since 13 is not a multiple of 4.

So, since 2016 is a multiple of 4, the solution is 2016/2 which is 1008. But, since 1008 is not a perfect square, he is the only solution.