Odd-length permutation forces an even product

Let $n$ be odd and let $A_1,A_2,\dots,A_n$ be a rearrangement of $1,2,\dots,n$. Prove that $(A_1-1)(A_2-2)\cdots(A_n-n)$ is even.

The factors sum to zero. $\sum_i (A_i-i)=\sum A_i-\sum i=0$. If every factor were odd, the sum of an odd number $n$ of odd integers would be odd, not $0$. So some factor is even, and so is the product.