If $p<n$ then it must be that $p \mid V$. However if $p \ge n$ then it can be arranged that $p \nmid V$. If you set $a_i=2^m(i-1)+1$ then no prime greater than $n-1$ divides $V$. You could replace $2^m$ by $(n-1)!$ or anything else with all divisors less than $n$.
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