Let's Bring It Back

If we want a perfect square, we must have even amount of all prime divisors. In $1! 2! 3! \cdots 2015! 2016!$ , the occurence of odd numbers ( not including its multiples) are always even, while occurences of even numbers ( not including its multiples) are always odd, so we don't have to care about the odd numbers. For example, the occurence of $3$ in all factorials ( $3!$ to $2016!$ ) is $2014$ times, while the occurence of $4$ in all factorials ( $4!$ to $2016!$ ) is $2013$ . The occurence of $2$ including its multiples in the even numbers is $1+2+3+\cdots+1008=508536$ , which is even, so we don't have to care about 2. Therefore we only want the even numbers divided by $2$ , which would be $1, 2, 3, \cdots, 1008$ , to have even occurences. Then we can divide the whole number by $1008!$ so that $1, 2, 3, \cdots, 1008$ would be cancled and have 0 occurences. Hence $x=1008$ .