How do I show that a prime that is less than $n$, is not a prime factor of $n$?

Question

I am studying the proof of the statement

there are infinitely many primes congruent to $1 \bmod 4$

Here is a snippet of the proof:

we first define $a$: $$a = (p_1p_2\cdots p_\ell)+1,\qquad p_i \in \mathbb P$$ but by unique prime factorization theorem, a can also be written as: $$a = q_1q_2\cdots q_n, \qquad q_i \in \mathbb P$$ we know $\{q_1,q_2,\ldots,q_n\}\cap\{p_1,p_2,\ldots,p_\ell\} = \emptyset$

How do I prove the last sentence? I was thinking using contradiction...assume that $p_i$ is in both set $$p_ix+1 = a = p_iy: x,y \in \Bbb Z$$ $$p_i(y-x) = 1$$ so that $p_i$ is not a prime, the only integer solution is $p_i = 1$ (it seems to me that if $a = (p_1p_2\cdots p_\ell)+2, p_i \in \mathbb P$, this proof will fail)

I didn't use any knowledge in math to prove it...is there a better proof?

Answer 1

$$ (p_1\cdots p_\ell) + 1 \overset{\large\text{?}} = q_1 \cdots q_n $$ $$ (p_1\cdots p_\ell) - (q_1 \cdots q_n) \overset{\large\text{?}} = 1 \tag 1 $$ What would happen if $p_1 = q_1$? Then $(1)$ would become $$ p_1 \Big( (p_2 \cdots p_\ell) - (q_2\cdots q_n) \Big) = 1. $$ That would mean $1$ is divisible by the prime number $p_1$. And we get the same conclusion if $p_1 = q_6$ or $p_8 = q_5$, etc. The number $1$ cannot be divisible by a primes, so we have to conclude $p_1$ is not in the intersection of those two sets. Nor is $p_2$, for the same reason, etc.

Answer 2

Let's say $p_j$ divides $a$. Then: $a = kp_j = \prod p_i + 1 \Rightarrow p_j\cdot(k- \underset{{i \neq j}}\prod p_i) = 1$. This means that $p_j$ divides $1$, therefore $p_j = 1$, which is a contradiction because $p_j$ is prime.

Therefore the two sets have to be disjoint.