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?

Let us assume the product is odd.

Then each of the factor of product must be odd, too.

Since sum of an odd number of odd integers is odd, the sum of the factors must be odd too.

But we've

$(a_1-1)+(a_2-2)+ \cdots + (a_n-n) = \color{red}0$

Which is even, a contradiction to our assumption.

Hence, the product is even.

As $n$ is odd then you have $\frac{n+1}{2}$ odd values of $a_i$ and $\frac{n-1}{2}$ even values of $a_i$. To get an odd product you need every odd value to be paired up with an even value so that their difference is odd. However due to the fact that there is one more odd number than even number you will always get at least one odd number paired with an odd number giving an even value in the product so the product is even.

The product is always even. To obtain an odd product you need that $a_k-k$ is always odd that is $a_k$ and $k$ should have different parity. But this is impossible because there are $(n+1)/2$ odd numbers and $(n-1)/2$ even numbers and $(n+1)/2>(n-1)/2$.

For the product of $2m+1$ integers (the differences) to be even, it is sufficient that one integer be even.

In $\{1, \cdots,2m+1\}$ there are $m$ even integers and $m+1$ odd.

By the "Pigeon-hole" principle you cannot arrange all the $m+1$ odd-pigeons into $m$ even-holes: one shall go into a odd cell, thus making an even difference.