Proof of the infinitude of primes Show that the integer $Q_n=n!+1$, where n is a positive integer, has a prime divisor greather han n. Conclude that there are infinity many primes.
Answer is given by:

Why is the following true: 
1)  $p|n!\Rightarrow p|(Q_n-n!)$
2) We came true a contradiction, but why is $p_1$ a prime divisor of $Q_1$ and $p_2$ a prime divisor of $Q_{p_1}$?
 A: If $p\le n$ we have that $p$ appears among $n$, $n-1$, $n-2$ and so on, so clearly $p\mid n!$.
If an integer $k$ divides the integers $a$ and $b$, then it divides $a-b$. Indeed, if $a=kx$ and $b=ky$, then $a-b=k(x-y)$.
This should explain the part in green: if $p\mid Q_n$ and $p\le n$, then $p\mid n!$ and so $p\mid(Q_n-n!)$. But $Q_n-n!=1$, so this is impossible. Hence $p>n$.
The argument shows that, given an integer $n$, a prime divisor of $Q_n$ is greater than $n$. Since $p_2$ is a prime divisor of $Q_{p_1}$, we have $p_2>p_1$. In general, $p_{k+1}$ is a prime divisor of $Q_{p_k}$, so $p_{k+1}>p_k$.
However, there's a better explanation: the set of prime numbers is unbounded (no $n$ can be an upper bound), hence it is infinite.
A: Because $n! = n\cdot (n-1) \cdot \ldots \cdot 1$ every integer which is less or equal to $n$ must divide $n!$. Since $Q_n = n!+1 > 1$ it has a prime divisor $p$. Now if $p \leq n$ then $p | n!$ and because also $p | Q_n$ (as it was a prime divisor of it) it must divide the difference $Q_n -n! = (n!+1)-n! =1$. But this can't be the case as $p>1$.
The second is true by construction as the prime divisors of $Q_n$ are greater than $n$ by the argument above. 
