Okay. So we wanna show that there are infinitely many prime numbers. Let's assume that there are only a finity amount of prime numbers which we'll label as p1, p2, ..., pn.
Now let's consider some number, N, which is obtained by multiplying all prime numbers and adding 1:
N = p1p2 ... pn + 1.
N must have a prime factor. This prime factor must be one of p1, p2, ..., pn, because they are all the prime numbers. But this causes a contradiction since N leaves a remainder of 1 when divided by any of the prime numbers.
Conclusion: there are infinitely many prime numbers.
Now let's go to our problem: we want to show that pn+1 ≤ p1p2 ... pn + 1.
Let's assume it is false, so: pn+1 ≥ p1p2 ... pn + 1.
Now consider the number N ≥ p1p2 ... pn + 1 where N = pn+1.
This number must have a prime factor, so we will have to divide by a prime number which is one of p1, p2, ..., pn.
But we can't do this because this leaves a remainder of 1, which leads to a contradiction.
Is it something like this? Sorry for the long post but this way I can summarise everything up for myself.
You're off to a good start. But the end is a little muddled. Call M=(p1p2...pn)+1. M has a prime factor, call it p, that is NOT one of the p1...pn. Good so far. Since p=pj for some j, it must be that j>n. Ok so far? Since you are listing the primes in ascending order it follows that p=pj>=pn+1. Since p is a factor of M, p<=M. So pn+1<=p<=M.