Symmetric group of composite degree implies every proper Hall subgroup is solvable

Statement
Suppose $$n$$ is a composite natural number. In other words, $$n > 1$$ and $$n$$ is not prime. Let $$S_n$$ denote the symmetric group of degree $$n$$, i.e., the symmetric group on $$n$$ letters. Then, any proper Hall subgroup of $$S_n$$ is solvable. In particular, $$S_n$$ is a group in which every proper Hall subgroup is solvable.

Facts used

 * Nonsolvable proper Hall subgroup of symmetric group is contained in symmetric group on subset of size one less