Prove that n!+1 contains a prime factor greater than n and use this to prove that there are infinte many primes

Question

Prove that $n!+1$ contains a prime factor greater than $n$ and use this to prove that there are infinitely many primes.

I said assume that $n!+1$ contains a prime $p$ which is less than or equal to $n$. So $p \leq n$, therefore $p \mid n!+1$, so $p$ has to divide $n!$ and $p \mid 1$. However since $p$ can't divide 1, we contradicted ourselves, so $n!+1$ must contain a prime greater than 1.

I don't know how to prove that there are infinitely many primes, and I'm not quite sure that my solution I gave above is correct.

Answer 1

This is pretty much like Euclid's original argument. Suppose that $n!+1$ has all prime factors less than $n$. However, then such primes would divide $n!$ since $n!$ is the product of all natural numbers less than. Therefore, $n!+1 \equiv 0+1 \equiv 1$ for any such primes, contradiction.

Answer 2

The solution you gave is quite correct. But you'd rather say $p\mid n!+1$ and $p$ divides $n!$ (for $p\leq n$), so $p$ has to divide 1.

For the infinite primes part, just reason like this: were there a highest prime number, say $n$, you have just proved $n!+1$ has a prime factor greater than $n$, hence leading to contradiction.