State/There are infinitely many prime numbers

$$N = \left( \prod_{i=1}^k p_i \right) + 1.$$ Then, $N \geq 2$, and so $N$ has a nonempty factorization into prime numbers. There exists a nonzero natural number $t$, and prime numbers $q_1, \ldots, q_t$, such that: $$N = q_1 \cdot \cdots \cdot q_t.$$ Since $q_1$ is a prime number, and all prime numbers were listed in the finite sequence $p_1, \ldots, p_k$, we find that $q_1 = p_i$ for some $1 \leq i \leq k$. It follows that $N$ is a multiple of the prime number $p_i$.

But, from the formula defining $N$, one sees that $N - 1$ is a multiple of $p_i$. Since $1 = N - (N-1)$, the two out of three principle implies that $1$ is a multiple of $p_i$. Hence $1 \geq p_i$, since multiples of $p_i$ are larger than $p_i$. This contradicts the fact that $p_i$ is a prime number.