Please prove that $(p!+1)$ is a prime, where $p$ is a prime. I got Euler's proof for the Question:
Prove that there is no largest prime.
But one of the solutions on internet gives this proof by Archimedes:

Please help me to understand the proof.
 A: Suppose $p$ is the largest prime.
By the fundamental theorem of arithmetic, you can write $p! + 1$ as a product of primes, which are necessarily not larger than $p$, and thus are factors of $p!$.
But $p! + 1$ divided by any integer $q \le p$ gives remainder $1$, as
$$
p! + 1 = q \cdot (p \cdot \ldots \cdot (q+1) (q-1) \cdot \ldots 2 \cdot 1) + 1,
$$ and so $p! + 1$ is not divisible by $q$.
A: Note that $p!+1\gt p$. By assumption $p$ is the largest prime, so $p!+1$ cannot be prime. Hence it has a prime divisor $q|p!+1$. But $q\le p$, so $q|p!$. From this we conclude $q|p!+1-p!=1$, which is absurd. If the assumption that there is a largest prime leads to a contradiction, then the assumption must be false and there is no largest prime.
A: This is a proof by contradiction. If you suppose $p$ is the largest prime, then $p! + 1$ being strictly larger than $p$ cannot be prime and hence by Fundamental Theorem of ARithmetic, there is a prime $q$ that divides $p! + 1$ and since $q < p$, $q | p!$ and since it also divides $p! +1$, it divides $1$, which is a contradiction.
A: $p! = p(p-1)(p-2)\cdots 1$ which is divisible by $q$ for all $q \leq p$. So if $p! + 1$ is divisible by prime $q$ where $1 < q \leq p$ (which must be true if $p$ is the largest prime because every number has a unique prime factorization), then you can subtract $p!$ from $p! + 1$ and the answer must again be divisible by $q$, but the difference is $1$ and $q > 1$, which gives a contradiction. So there cannot be a largest prime $p$.
