, 2011

Let $p < a$ be a prime. If $p \nmid n$, the claim holds.
Suppose then that $n$ is not divisible by $p$. Then the numbers $2n + 1, 3n + 1, \dots, (p+1)n + 1$ give different remainders when divided by $p$, which are denoted, respectively, by $r_1, r_2, \dots, r_p$. Indeed, if $r_i = r_j$, then
$$p|((i+1)n + 1) - ((j+1)n + 1) = (i-j)n,$$
which implies that $i = j$. So each of the $p$ possible remainders appears exactly once. In particular, one of these remainders must be zero, meaning that the one of the factors in the product is divisible by $p$, which finishes the problem.