by contradiction, assume finitely many primes $p_1, p_2,\cdots, p_k$. let $N = p_1p_2\cdots p_k + 1$. Note $N > 1$. Now, by the fundamental theorem of arithmetic, there exists a number $p_j$, where $j$ is an element of $(1,2,...,k)$ such that $N = P_j\times q$. So, $p_1p_2\cdots p_k + 1 = p_j*q$. So, $1 = p_j*(q - p_1p_2p_3\cdots p_k)$. So $p_j|1$, but now we reached a contradiction, $p_j < 1$ but $p_j > 1$.
Can you please explain two parts
1) How does it become $1 = p_j(q-p_1p_2p_3\cdots p_k)$
2) the contradiction, why does it mean $p_j <1$, couldn't $p_j = 1$(but I guess it's still not a prime though)