, 2015

Let $a_1 = 2$, $a_2 = 3$. For $i = 3, \dots, n-1$, let $a_i = a_1 a_2 \cdots a_{i-1} + 1$. Let $a_n = a_1 a_2 \cdots a_{n-1} - 1$. Clearly, $a_1 a_2 \cdots a_{n-1} \equiv 1 \pmod{a_n}$. Also $a_{i+1} \equiv a_{i+2} \equiv \cdots \equiv a_{n-1} \equiv 1 \pmod{a_i}$. For $i = 1, \dots, n-1$, we have
$$a_1 a_2 \cdots \hat{a}_i \cdots a_n = (a_1 \cdots a_{i-1})(a_{i+1} \cdots a_{n-1})a_n \equiv (-1)(-1)(-1) \equiv 1 \pmod{a_i}.$$