About the parity of the product $(a_1-1)(a_2-2)\cdots(a_n-n)$ An exercise from Chapter 20 of "How to Think Like a Mathematician" by Kevin Houston:

Let $n$ be an odd positive integer. Let $(a_1,a_2,\dots,a_n)$ be an arbitrary arrangement (i.e., permutation) of $(1,2,\dots,n)$. Prove that the product $(a_1-1)(a_2-2)\cdots(a_n-n)$ is even.

For example, for $n = 3$, we can have $(a_1, a_2, \dots, a_n) = (3, 1, 2)$, so this yields that $(3-1)(2-1)(1-2)$ is even.
Would the following be considered a full solution?
Each even an has to be paired with a ($a_n$ -odd) in order for each even an to become odd. However after pairing them we have one -odd left as there is one more odd than even in the set $(1,2,\dots,n)$ where $n$ is odd. This must be paired with a odd number in a bracket. As odd-odd is even (the proof of this is trivial), then one of the bracket must be even so the product is even.
Would THIS solution be considered full?

It suffices to show that one of the terms $a_k-k$ is even.
For that to fail, we would need each $a_k$ be of a parity distinct to that of $k$.
Since $n$ is odd, in $\{1,2,\dots, n\}$ there are $\frac{n+1}2$ odd numbers and $\frac{n-1}2$ even numbers.
But in order for all the $a_k$ to be a distinct parity that of $k$, we'd need the $a_k$ to have $\frac{n-1}2$ odd numbers and $\frac{n+1}2$ even numbers.
