Trying to understand Euclid's prime number proof I am not a mathematician, so I don't really understand one specific of Euclid's proof. I am, however, trying to learn about prime numbers. Here's my question:
Why does Euclid use the set $P=p_1*p_2...*p_n$ instead of $R=p_1*p_2...*p_{(n-1)}$?
The reason why I am asking is because, and correct me if I'm wrong, as you move towards infinity, the distance between every prime number should logically get larger. For me, this brings up several questions. Assume $Z= 1$, $2$ or $3$. These three prime numbers are the only known primes in which the statement R < $p_n$ holds true. After the first three prime numbers, the subsequent primes follow the pattern of  $R > p_n$
With that being said, if the distance between each prime number logically increases as you move towards infinity, shouldn't there eventually be an instance in which $R = p_n$ or... $R < p_n$ again?
In my mind at least, I can't understand what happens to the prime numbers after $R = p_n$ then passes into $R < p_n$. Does anything change with the prime numbers?
You see, with a different subset of numbers, I came to a completely different answer - that, maybe, $R < p_n$ cannot exist past 1, 2 or 3, so there could really be a 'highest prime number.' Am I making any sense? Why did Euclid use set P instead of R?
Thanks!
 A: The average distance between primes increases as you get into larger and larger primes. But it increases slowly.
The product $p_1 p_2 p_3 \cdots p_{n-1},$ however, grows very quickly:
\begin{align}
2 \cdot 3 &= 6 \\
2 \cdot 3 \cdot 5 &= 30 \\
2 \cdot 3 \cdot 5 \cdot 7 &= 210 \\
2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 &= 2310 \\
2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 &= 30030 \\
\end{align}
By this point we're only up to $13$ as the largest prime on the left hand side, and yet the right hand side is already in the tens of thousands.
And it only gets more extreme as we keep multiplying by more primes.
So no, you would never get to a point where $R < p_n.$
On the contrary, the ratio $R/p_n$ keeps attaining larger and larger values.
By the way, there is no guarantee that by applying Euclid's argument to 
$p_1 p_2 p_3 \cdots p_{n-1}$ you will discover the prime $p_n.$
Yes, as a comment by lulu showed, when $n = 7,$ then $p_{n-1} = 17,$
and $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 +1 = 510511,$ 
which is divisible by the eighth prime, $19.$
But if we set $n=8$ then we have
$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 +1 = 9699691,$
whose prime factorization is
$9699691 = 347 \times 27953.$
That's right, by the time we get to $n=8$ the product 
$p_1 p_2 p_3 \cdots p_{n-1}$ is so enormous that the two prime factors of
$p_1 p_2 p_3 \cdots p_{n-1} + 1$ are both many places beyond $p_n$ in the list of primes.
But boiled down to its essence, Euclid's proof argues that there is no such thing as the finite product of all the primes.
By adding one to the alleged finite product of all the primes, we discover that it "missed" at least one prime;
so if the list of all primes were finite, it wouldn't be a list of all the primes.
To make this work, however, we have to look at the product of all of the
(allegedly finite number of) primes that exist.
If we only have a list of some of the primes that exist, it would hardly be surprising that the "new" prime exhibited by Euclid's argument would be one that we already knew existed but didn't bother to put in the list.
Whether we number the list of all primes from $1$ to $n$ or from $1$ to $n-1$ doesn't matter; what matters is the "all of the primes" part.
A: https://www.youtube.com/watch?v=ctC33JAV4FI here's a basic overview of the video using the modern way to demonstrate the proof. By the way  they probably use n as the index to go up to because n is easier to write. 


*

*Let's assume the set of primes is a finite set $\{p_1,p_2,p_3,...p_n\}$

*Let's say the the natural numbers are infinite.

*This means we can make a number in the list by multiplying all the primes together and adding one. ( $P=1+\prod_\limits{k=0}^n p_k$)

*Using the fact that no number greater than a  number divides into that number to give a natural number result we get that P can no longer divide by any of our prime set.

*P is either prime or it's composite. If it's prime, it's missing from our set (so we started with an incomplete set). If it's composite it's prime factors aren't in our set ( so we started with an incomplete set). This contradicts that our set of primes contained all the primes in either case. 

A: It doesn't matter whether you use  $n $ or $n-1$, the point is that you assume  finitely many primes and you get a contradiction. ..
