imc

Perfect squares have all of the powers in their prime factorization even. To evaluate $2!\cdot 3!\cdot 4!\cdot 5!...16!$ we get the following: $$\begin{gathered} (2^{15})\times (3^{14})\times ((2^2{)}^{13})\times (5^{12})\times ((2\times 3{)}^{11})\times (7^{10})\times ((2^3{)}^9)\times ((3^2{)}^8)\times \\ ((5\times 2{)}^7)\times (11^6)\times (((2^2)\times 3{)}^5)\times (13^4)\times ((7\times 2{)}^3)\times ((3\times 5{)}^2)\times ((2^4{)}^1) \end{gathered}$$. Taking all powers $\mathrm{mod}2$ we get: $(2^1)\times ((2\times 3{)}^1)\times (2^1)\times ((5\times 2{)}^1)\times (3^1)\times ((7\times 2{)}^1)$ Simplifying again, we finally get: $(2^1)\times (5^1)\times (7^1)$ To make all the powers left even, we need to multiply by $(2\times 5\times 7)$ which is $\boxed{70}$.