1. The problem statement, all variables and given/known data Prove that n! for n>1 cannot be a square or cube or any other power of an integer. Hint: There is always a prime between n/2 and n if n>3 3. The attempt at a solution Lets assume for contradiction that [itex] n!=x^r [/itex] where x and r are natural numbers and n>3 , so there is some prime p that is between n/2 and n. well this occurs once in prime factorization of n!. Now x has some prime factorization and in order for [itex] x^r [/itex] to be equal to n! the prime p must be in x's prime factorization, but if r>1 then p will have r copies in [itex] x^r [/itex] so this is a contradiction because numbers have unique prime factorizations so n! is never a power of an integer for r>1.