$S_n$ is a set of all possible permutations of length $n$ here. I think solving it with decrement is too hard and inversions should be used instead. So, it's obvious that $Id$ has zero inversions and 1->n, 2->n-2,...,n->1 has $ \binom{n}{2}$ inversions and it's max possible. Then I tried to find permutation with $\binom{n}{2}-1$, $ \binom{n}{2}-2$ number of inversions so that i could compute desirable product but kinda failed.
Assume $n\geq2$, and fix an arbitrary transposition $s$ (transpositions swap two elements, they are odd permutations). There is a bijection between the even permutations of length $n$ and the odd permutations of length $n$ given by taking any permutation $t$, and returning the permutation given by first applying $s$, then applying $t$. In other words, $t\mapsto ts$. We see it is bijective because it is its own inverse. This proves that there are equally many odd and even permutations.
Since there are equally many even and odd permutations, there are $\frac{n!}2$ odd permutations, meaning the given product is equal to $1$ for $n\geq4$, and $-1$ for $n=2,3$.