Define $\mathbb{N}_n=\{1,...,n\}$
The pigeon hole principle as I know is that if $n>m$. Then for every map
$f:\mathbb{N}_n\rightarrow\mathbb{N}_m$ there exists two distinct $n_1,n_2$ such that $f(n_1)=f(n_2)$
The excercise is if $a_1,...,a_n$ is an arrangement of the numbers $1,...,n$ then
$(a_1-1)(a_2-2)….(a_n-n)$ is even.
If the prduct would not be even then every factor would have to be uneven. That means $a_1,a_3,..a_n\in\{2,4,...,n-1\}$.
The only map that I have is the arrangement $f:\{1,...,n\}\rightarrow\{1,...,n\}$
How can I construct a contradiction with the pigeon hole principle as stated above now?
EDIT:
$n$ is uneven
To elaborate on my comment:
In order for this product to be odd, we need every factor in it to be odd. Thus each number $i$ must be associated with another element of the set of the opposite parity. Evens must be associated with odds, and odds with evens. In particular, we need there to be as many evens as odds.
But: if $n$ is odd then the list $\{1,\cdots, n\}$ contains more odds than evens. Thus it is not possible that each odd $i$ is associated with a different even.
Note that, if $n$ is even, the claim is false. We can just shift each element one unit to the right (cyclicly) to get an odd product.