You can't prove it because it is false.

$\dbinom{6}{2} = \dfrac{6!}{(6 - 2)! * 2!} = \dfrac{6 * 5 * 4!}{4! * 2!} = \dfrac{6 * 5}{2} = 15.$

Last I looked, 3 is prime, 15 is divisible by 3, and 6 is not a power of 3 though 6 is an integer multiple of 3.

EDIT: upon reflection, 5 is prime, 15 is divisible by 5, and 6 is not even an integer multiple of 5.