Backward direction – Wilson’s Theorem – p is prime $\iff (p-1)!\equiv-1(mod\ p) $. 
(1) How can you preconceive to prove by contradiction?

Prove by contradiction. Suppose $n$ is composite. This means there exists a divisor $d|n$ such that $1<d<n$. We are given that $(n-1)!\equiv-1(mod\ n)$ which means $n|[(n-1)!+1]$. Since $d|n$ so
$d|[(n-1)!+1]$ which gives $dk=\color{seagreen}{ (n-1)! }+1$ for some integer $k$.
From the first line, and independently from the above paragraph, we have $1 < d < n$. Therefore $d|(n-1) ! \implies \color{seagreen}{dm=(n-1)}$ ! for some integer $m$.

(2) How can you preconceive to consider $1 < d < n$ alone and separately, in the middle of the proof?
(3) Why $d|(n - 1)!$ ? By reason of $1 < d < n \iff 2 \le d \le n - 1$?

Substitute $dm$ into $dk$, $dk=dm+1 \iff d(k-m)=1 \iff d$ divides 1.
Origin — p4 —  better than Jones p82 Question 4.20
 A: 
(1) How can you preconceive to prove by contradiction?

Consider a few examples.  $3! = 6$, $4! = 24$, $5! = 120$.  These are highly divisible numbers, and in particular $(n-1)!$ contains every number up to $n-1$.  You therefore expect that if $n$ is not prime, then $(n-1)!$ will be divisible by $n$.  Hence, prove by contrapositive, or alternately proof contradiction will do the trick.  (Note: $n = 4$ is the only exception where $(n-1)!$ is not divisible by $n$, but it still has some of the same factors.)
Beyond that, note that proof by contradiction, in general, is easier than both a normal proof and a proof by contrapositive, because you get to assume twice as many things.  Thus, when stuck, always try a proof by contradiction.

(2) How can you preconceive to consider $1 < d < n$ alone and separately, in the middle of the proof? 

You essentially want to show that $(n-1)!$ has some of the same factors as $n$.
Thus naturally you find some $d$ which should be in the expansion of $(n-1)!$.
So you need $1 < d < n$.

(3) Why $d|(n - 1)!$ ? By reason of $1 < d < n \iff 2 \le d \le n - 1$?

Any positive integer less than $n$ is included in the expansion
$$
(n-1)! = (n-1)(n-2) \cdots (2)(1).
$$
In particular for $d$ to be in this expansion, we need $2 \le d \le n-1$.
So yes, that is the correct reason.
