Wilson's Theorem textbook proof question 
I'm trying to understand this proof from Stein's Elementary Number Theory, and I understand the pairing of inverses but not the other direction. I have two questions:
$1).$ When the proof says, $l$ a prime divisor of $p$, and so $l<p$ and $l \mid (p-1)!$, is this because $l \mid p \Rightarrow l \mid p!=p*(p-1)!$ and so $l \mid (p-1)!$ (by Euclid)?
$2.)$ Why is it that $p \mid ((p-1)!+1)$?
 A: $(2)\ \ $ $(p-1)!\equiv -1 \pmod p\,\Rightarrow\, p\mid (p-1)!+1$
$(1)\quad \color{#c00}\ell < p\ \Rightarrow\, (p\!-\!1)! = (p\!-\!1)(p\!-\!2)\cdots \color{#c00}\ell \cdots 2\cdot 1\,\Rightarrow\, \color{#c00}\ell\mid (p\!-\!1)!$
Since also $\,\color{}\ell\mid p\mid 1\!+(p-1)!$ we infer $\,\ell$ divides their difference $= 1,\,$ contradiction.
A: 1) That argument would work if $l$ didn't divide $p$ (as $l | ab$ implies $l | a$ or $l|b$, since $l$ is prime.) Instead, the point here is that $l\not = p$ is a divisor of $p$, and so must be one of $1, \dots, p - 1$. Thus it divides $1.2.\cdots.(p-1) = (p - 1)!$.
2) It was proved above (just after the second set equation) that $(p - 1)!\equiv -1\pmod{p}$, which just means that $p$ divides $(p - 1)! + 1$.
A: Regarding 1: $l \mid (p-1)!$ because $(p-1)!$ is the product of the positive integers less than $p$, so every positive integer less than $p$ divides $(p-1)!$.
Regarding 2: $a \equiv b \pmod n \iff n \mid (b - a)$
A: For (1), the key is that every number less than or equal to $p-1$ divides $(p-1)!$. Since $p$ is assumed to be composite at that line of the proof, the fact that $l$ is prime and divides $p$ implies that $l \leq p-1$. It's not for the reason you suggest: $l$ divides $p$, so there is no reason it must divide $(p-1)!$ just because it divides $p!$.
For (2), note that $(p-1)! \equiv -1 (\mod p)$. 
