What is the number of Sylow p subgroups in $S_p$?

I am reading the Wikipedia article entitiled Sylow theorems. This short segment of the article reads:

Part of Wilson's theorem states that $(p-1)!$ is congruent to $-1$ (mod $p$) for every prime $p$. One may easily prove this theorem by Sylow's third theorem. Indeed, observe that the number $n_p$ of Sylow's $p$-subgroups in the symmetric group $S_p$ is $(p-2)!$. On the other hand, $n_p ≡ 1$ mod p. Hence, $(p-2)! ≡ 1$ mod $p$. So, $(p-1)! ≡ -1$ mod $p$.

I do not understand why the number of Sylow p-subgroups of $S_p$ is $(p-2)! $

I am taking a course in Group Theory and I have studied everything in Dummit and Foote up to Sylow's theorem. I have also had an introductory course in number theory and am familiar with basic combinatorics.

The elements of order $p$ consist of a $p$-cycle of the form $(1,a_2,a_3,\ldots,a_p)$, where $a_2,a_3,\ldots,a_p$ is a permutation of $2,3,\ldots,p$. So there are exactly $(p-1)!$ elements of order $p$.

Now each subgroup of order $p$ contains $p-1$ elements of order $p$ (i.e. its non-identity elements), and the intersection of any two such subgroups is trivial, so the total number of subgroups of order $p$ is $(p-1)!/(p-1) = (p-2)!$.