Infinite prime proof using factorial plus one or product of primes plus one?

You can use either one. The product of all the primes up to $p$ is called the primorial and sometimes written $p\#$. We have $p\#$ divides $p!$ because $p!$ includes all the primes less than $p$ as factors along with other numbers. The statement that any prime dividing $p\#+1$ or $p!+1$ must be greater than $p$ goes through and provides a prime greater than $p$ as required.

The usual proof works with the product of the primes up to $p$, but this argument with the factorial $p!$ works just as well.

The factorial and the primorial are different things. Given a positive integer $n$, the factorial $n!$ is divisible by each integer from $1$ to $n$, whether it's prime or not. e.g., $5! = 1 \times 2 \times 3 \times 4 \times 5 = 120.$

What is $5$ primorial? There is some disagreement: it could be $2 \times 3 \times 5 = 30$, or it could be $2 \times 3 \times 5 \times 7 \times 11 = 2310$ (the product of the first five primes).

But if we are agreed on the definitions, we can use either factorials or primorials. The important thing is that we have some number divisible by lots of distinct primes, and one which is divisible by none of those (add or subtract $1$ from the former to obtain the latter).

The factorials have a slight disadvantage in that they get larger quicker. Compare $29! = 8841761993739701954543616000000$ to the product of the primes from $2$ to $29$, which is $6469693230$.

Now notice that $6469693231 = 331 \times 571 \times 34231$, while the least prime factor of $8841761993739701954543616000001$ is $14557$.