Symmetric group on finite set has Sylow subgroup of prime order

Statement
Suppose $$n$$ is a natural number greater than $$1$$. Let $$S_n$$ be the fact about::symmetric group of degree $$n$$. Then, there exists a prime $$p$$ such that $$S_n$$ has a $$p$$-fact about::Sylow subgroup of order $$p$$, i.e., it is a fact about::group of prime order.

Facts used

 * 1) Bertrand's postulate: The version we use states that for any $$n \ge 2$$, there exists a prime $$p$$ such that $$p \le n < 2p$$.
 * 2) uses::Sylow subgroups exist

Proof
The symmetric group of degree $$n$$ has order $$n!$$. By fact (1), there exists a prime $$p$$ such that $$n/2 < p \le n$$. Thus, the largest power of $$p$$ dividing $$n!$$ is $$p$$. In particular, this means that the order of the $$p$$-Sylow subgroup (which exists by fact (2)) is $$p$$.