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?

At first, let us assume that $(a_n)$ has no fixed points, so:
$$a_k\neq k,\ \forall k=1,2,\dots,n$$
since, if it has, the problem is trivial - the product is equal to zero.
Now, suppose that $$p=(a_1-1)(a_2-2)\dots(a_n-n)$$
is odd, so, $$2\not|p\Rightarrow 2\not|(a_k-k),\ \forall k=1,2,\dots,n$$
since, if $2$ divided some factor, $p$ would be even. But, then, if $m$ is any even number form $\{1,2,\dots,n\}$, it means that $a_m$ is odd - else, $a_m-m$ would be even, which is not possible. So, $$\{2,4,\dots,n-1\}\overset{a}{\mapsto}\{1,3,5,\dots,n\}$$
which is not possible, since the first set has exactly $\frac{n-1}{2}$ elements and the second $\frac{n+1}{2}$, since $n$ is odd.
