Let $N$ be the product of the first $m$ primes and $M$ be the product of the first $n-m$ primes. Could someone tell me if this proof is correct?
Suppose there are finitely many primes: $p_1,p_2,\cdots,p_n$ primes.  Let $m<n$.  Then let $N=p_1p_2p_3\cdots p_m$ and $M=p_{m+1}\cdots p_{n-1}p_n$.  Let $P=M+N$.  First, $P$ cannot be prime or else we have a prime bigger than $p_n$.  So $P$ is composite, hence there exists a $p_i$ such that $p_i \mid P$.  But if $p_i\mid P$, then $p_i \mid M$ or $p_i \mid N$.  So $p_i \not\mid P$.  Hence the $P$ has a prime factor not on the list.

How about this: Suppose there are finitely many primes say $p_1,p_2,\dots,p_n$. Let $N=p_1p_2\cdots p_m$, $M=p_{m+1}\cdots p_{n-1}p_n$ and $P=M+N$, where $m<n$. Since  any prime number divides exactly one of $N$ and $M$, so no prime divides $P$, i.e. $P=1$, a contradiction.
