By the Chinese remainder Theorem, for every $k=2,3,\dots,n$ there exists $b_k$ so that
$$b_k \equiv 0 \pmod{(k-1)}, \quad b_k \equiv k \pmod{n}.$$
Let $a_1 = 1$ and for $k = 2, \dots, n$, $a_k$ is the remainder when $b_k/(k-1)$ is divided by $n$. Since $b_n \equiv 0 \pmod{n}$, we have $a_n = 0$. Also if $a_i = a_j$, then $b_i/(i-1) \equiv b_j/(j-1) \pmod{n}$, i.e., $i(j-1) \equiv j(i-1) \pmod{n}$. Since $n$ is prime, $i = j$. Thus $a_1, \dots, a_n$ are distinct. Now
$$a_1 a_2 \dots a_k \equiv \frac{b_2 \dots b_k}{(k-1)!} \equiv k \pmod{n}.$$