- #1
cragar
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Homework Statement
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
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.