Pick any prime $p_i$ from our list $L = \{p_1, p_2, \dots, p_n\}$. In the product $P = p_1 p_2 \cdots p_n$, we can factor out $p_i$ and write $P = p_i \cdot K$, where $K = \prod_{j \neq i} p_j$ is the product of the remaining primes in the list (or $K = 1$ if $n = 1$). Substituting this into the definition of $N$ gives $N = p_i \cdot K + 1$.

This equation says that when $N$ is divided by $p_i$, the quotient is the integer $K$ and the remainder is $1$. If $p_i$ were to divide $N$ evenly ($p_i \mid N$), then since $p_i$ also divides $p_i \cdot K$, it would have to divide their difference $N - p_i \cdot K = 1$. In Euclid's words in Book IX, Proposition 20, the prime "measures the remainder, the unit $DF$, which is absurd," because every prime satisfies $p_i \ge 2$ and no integer $\ge 2$ can divide $1$. Thus $p_i \nmid N$ for every $p_i \in L$.