7! = 5040 = 2^4 * 3^2 * 5 * 7
Sqrt{2^4 * 3^2 * 5 * 7]= 2^2 * 3 * sqrt(5 *7) = 12 sqrt(35) - which is what you want.
+62
To calculate the square root of 7! where n! stands for n*(n-1)(n-2)...2*1, we can use the factorial property:
7! = 7654321
sqrt(7!) = sqrt(7654321) = sqrt(7) * sqrt(6) * sqrt(5) * sqrt(4) * sqrt(3) * sqrt(2) * sqrt(1)
sqrt(7) * sqrt(6) * sqrt(5) * sqrt(4) * sqrt(3) * sqrt(2) * sqrt(1) = 7^(1/2) * 6^(1/2) * 5^(1/2) * 4^(1/2) * 3^(1/2) * 2^(1/2) * 1
Using this information to jumpstart your question, try simplifying the last expression.
