Understanding the proof for showing that there are infinitely many $n\in\mathbb{Z}^+$ such that $n!+1$ is divisible by at least two distinct primes There is another post pertaining to this question; however, the other post does not address some specifics that concern me in the proof.
Proof: Let $n=p-1, p \geq 7$. Then $p|(p-1)!+1$. We will show that $(p-1)!+1=p^n$ has no solutions.
Suppose not. Suppose $(p-1)! + 1 = p^n$ has a solution for $p\geq 7$. Then $(p-1)! =p^n-1  = (p-1)(p^{n-1}+\cdots+p+1)$. So $(p-2)!\equiv n (\mod p-1)$.
Now suppose that $(p-1)|n$. Then $p-1\leq n$. So $(p-1)!+1<p^{p-1}\leq p^n$. Hence, there are no solutions.
It remains to show that $(p-1)|(p-2)!$. But $p-1=2m$ means $2m|(p-2)!$. QED

Questions:

*

*My understanding is that at the start of the proof, Wilson's theorem is used to show that at least one prime divides $n!+1$.


*Why do we want to show that $(p-1)!+1=p^n$ has no solutions? The way I see it is that $p^n$ is composite, and we are basically saying that $p^n\nmid (p-1)!+1$, so no composite number divides $n! +1$. But how and where in the proof do we actually show that another prime $q \neq p$ divides $n!+1$?

Thanks in advance.
 A: You want to show that at least two primes divide $n!+1$. 
What is the exact opposite of this? 
"$n!+1$ has only one prime divisor, i.e. $n!+1$ is a power of some prime number."
Now, suppose $(p-1)!+1$ is power of a prime number for some $p\ge 7$. However, we already know that $p|(p-1)!+1$, so, if $(p-1)!+1$ is power of a prime number, it needs to be $p^n$, for some $p$. Your proof shows that this is not possible, although you should've specified that as $p\ge 7$, $p-1>4$, so, $m>2$. 
Thus, $(p-1)!+1$ has at least two different prime divisors for all prime number $p\ge7$. As there are infinitely many prime numbers greater than $7$, we're done.
A: Suppose $ p $ is a prime and $ p \geq 7\ $. I think that Lucas' proof  (above) only shows that $  (p-1)!+ 1 \neq p^{p-1} $. We prove that
 $  (p-1)!+ 1 \neq p^{k}$, for $  k \geq 1 $. Then, since, by Wilson's Theorem,  $ p \mid (p-1)!+1 $, it follows that at least two distinct primes divide $ (p-1)!+1$. Lower-case Latin letters denote non-negative integers.
Lemma 1. If $ $$ n\geq 7 $ and $n$ is odd, then, $ n-1 \mid (n-2)! $.
Proof. Let $t = \frac{n-1}{2} \in Z $. Then, $2<{t} \leq {n-2}$. Lemma 1 follows, since $n-1=2t\blacksquare$
Lemma 2. If $ $ $ 2 \leq t \leq n$, then, $n^t-n^{t-1} \geq t!$
Proof. If $n \geq 2$, then $n^2-n = (n - \frac{1}{2})^2-\frac{1}{4} \geq (2 - \frac{1}{2})^2-\frac{1}{4} = 2!$  . Thus, Lemma 2 holds for
$t=2$. Suppose Lemma 2 held for some $t$, where $2 \leq t < n$. Then,
$n^{t+1} - n^t =n(n^t-n^{t-1}) \geq n(t!) \geq (t+1)!$. So, Lemma 2 holds by  induction on $t\blacksquare$
Suppose $(p-1)!+1=p^k.$ Clearly, $k>0$. Also, $p \equiv 1$(Mod $ p-1$). Therefore,
(1)  $p^t \equiv 1^t = 1$ (Mod $p-1$), for $t \geq 0$. 
$(p-1)! = p^{k} - 1= (p-1) \sum_{t=0}^{k-1}p^{t}$. Dividing this by $p-1$ and using (1) yields
$(p - 2)! = \sum_{t=0}^{k-1}p^{t} \equiv \sum_{t=0}^{k-1}1 = k$(Mod $p-1$). So, $p -1\mid k$, by Lemma 1. Hence, $k \geq p-1$,
since $k > 0$. Therefore, by Lemma 2,with $n = p$ and $t = p-1$,
$(p-1)!+1=p^k \geq p^{p-1} \geq (p-1)! + p^{p-2} > (p-1)! + 1$, a contradiction$\blacksquare$
