As Eric Naslund makes clear, his proof can be adjusted to deal with the $2$, and most reasonable variants that involve a fixed number $n$ of terms. The following is a very elementary argument for a much weaker result.
Let $q_1,q_2,\dots, q_{n}$ be a non-empty list of odd primes such that the list contains an even number (possibly $0$) of primes of the form $6k-1$. Then $2q_1q_2\cdots q_{n}$ is congruent to $2$ modulo $3$, and hence $2q_1q_2\cdots q_{n}+1$ is divisible by $3$. Since this number is greater than $3$, it is composite.
As a variant, let $p_0=2$, $p_1=3$, $p_2=5$, and so on. Let $a_n=p_0p_2p_3\cdots p_n+1$. (Note that we are skipping the prime $3$.) Then the sequence $(a_n)$ contains infinitely many composite numbers.
The proof uses nothing more than Euclid's Theorem.
It can be proved, by a mild variant of Euclid's proof, that there are infinitely many primes of the form $6k-1$. (There are also infinitely many of the form $6k+1$, but the proof is a little harder.)
From the fact that there are infinitely many primes of the form $6k-1$, we can, by taking an even number of them, and if necessary adding $5$ to the list, produce, for any fixed $n>1$, infinitely many $n$-tuples $(q_1,q_2,\dots,q_n)$ of distinct primes such that $2q_1q_2\cdots q_n +1$ is composite. The fact that there are infinitely many primes of the form $6k+1$ gives us in addition the result for $n=1$.
Comment: The numbers $a_n=p_0p_2p_3\cdots p_n+1$ mentioned in the solution are close relatives of $b_n=p_0p_1p_2\cdots p_n+1$. But while the proof that the sequence $(a_n)$ has infinitely many composites was elementary, at this time it is not known whether the sequence $(b_n)$ has infinitely many composites.