Euclid's proof (Variations)

Question

I have seen the following proof of Euclid's theorem:

Euclid's proof (Variations)

In the variant, they don't say it is $n$. Should I assume that $n$ is a natural number greater than $1% or should I assume that $n$ is any prime number?

Also found the following demo:

Given a number $n$, let's consider n! It is true that everything number less than or equal to $n$ divides $n!$, then no number less than or equal to that $n$ divides $n! + 1$. Consequently a prime divisor of $n! + 1$ must be greater than $n$. Therefore above every number $n$ there is always a number prime. This implies that there are infinitely many primes.

Again, in this new proof I've cited, should I assume that $n$ is a number greater than $1$ or a prime number?

Answer 1

It's proof by contradiction assuming that,

There exists an integer n, such that n is the largest prime.

See the book quoted from the Euclid-Variation, Bostock, Linda; Chandler, Suzanne; Rourke, C. (2014-11-01). Further Pure Mathematics. Nelson Thornes. p. 168. ISBN 9780859501033.

While wiki page Euclid-Variation doesn't give a full picture, but only mention that

The factorial n! of a positive integer n is divisible by every integer from 2 to n, as it is the product of all of them.

It would be better to have the above contradiction statement in the book, thus n starts from 2.

Answer 2

Suppose $m$ is the largest prime.

Let $p_1, p_2, ..., p_n$ be primes less than $m$.

What is $q = (p_1 \cdot p_2 \cdot ... \cdot p_n \cdot m) + 1$? Is it not prime and is it not greater than $m$, the alleged "largest" prime?

Remember, it must be that for $q$ to be composite, $q$ is a unique product of nonzero powers of $p_1, p_2, ..., p_n$ i.e. it's at a minimum divisible by $p_1, p_2, ..., p_n, m$, but that isn't true (because of $+ 1$).