Factorization of integers - why does it suffice to consider squarefree instances?

Question

I sat a lecture where a proposition is proven that states the following:

If computation of $(k!)_{k\in\mathbb{N}}$ is "easy", then integer numbers can be factored in non uniform polynomial time.

The proof picks a number $n$ and asserts right from the start that $n$ is square-free. My question is simply: Why does it suffice to consider square-free numbers?

Answer 1

Possibly, it might be because of the Moebius function:

$\mu(n)=0 \iff n\ $ is not square-free,

and $\mu$ can be calculated using the primitive $n$th roots of unity: see e.g. here.