Yes, I guess you could say that x*(a^n) = 8!.

I have the feeling that this question has been answered before but here goes.

1. a^n = 64 is actually telling you to look for powers of 2.
64 = 2^6 = 4^3 = 8^2 - as you can see, there are three possible values for a, so 1 is insufficient.

2. Now look at 8!. You could very easily do the prime factorization of this number:
8! = 1*2*3*4*5*6*7*8

8! = 2*3*(2^2)*5*(2*3)*7*(2^3)

8! = (2^9)*(3^2)*5*7.

As you can see, if a^6 is a divisor of 8!, then a can only be 2. 3^6 is not going to be a divisor of 8!, since there aren't enough 3's in the prime factorization. Since you get a single value for a, then 2 is sufficient.