Factorial Factors Reloaded

Note that $10!=2^8 \times 3^4 \times 5^2 \times 7$ . The key to this problem is to notice that k can be negative integer as a square of any integer is always positive. For $k^2$ to divide 10!, not more than twice the power of each term of the prime factorization of k should exceed the power of each term in the prime factorization of 10!. So the minimum possible value of k is $-(2^4 \times 3^2 \times 5)=-720$ .