Order of a Sylow $p$-subgroup of the symmetric group of order $p!$

Question

I have been struggling with this question because many times I saw the statement "order of Sylow $p$-subgroup of symmetric group of order $p!$ is $p$" being used in exercises and texts without argumentation.

I know that it has to be of form $p^k$, but why is it that we automatically know the order of Sylow $p$-subgroup of symmetric group of order $p!$ is $p$, i.e. $k=1$?

Does this state only for $n=p$ or more general?

Answer 1

$p$ cannot be a prime factor of any number smaller than $p$, so $p!$ has exactly one factor of $p$.

Answer 2

Result is true for any symmetric group of order $k!$ where $k=p,p+1,\dots,2p-1$