Prove that an integer n>1 is square-free iff n can be factored into a product of distinct primes.

My proof.

Assume $\displaystyle n=p_{1}p_{2}...p_{n} $ for p being distinct primes. Since all the factors are no equal to one another, n cannot be divided by a square of an integer.

conversely, assume n to be square free, then n cannot be divided by a square of an integer...