Let $n$ be odd and supposed $(x_1,x_2,...,x_n)$ is any permutation of $\{1,\ldots,n\}$. Prove that the product $(x_1-1)(x_2-2)\ldots(x_n-n)$ is even.

Question

Is the result necessarily true if $n$ is even? Give a proof or counterexample.

For the first part, we know $n$ is odd so we let $n = 2m + 1$. Then let $F = (x_{1},...,x_{n})$, and $S = \left\{a\, | \,F(a) \text{ is odd}\right\}$. Then we know that $|S| = m + 1$, and we know that there are only $m$ even integers in $\left\{1,...,n\right\}$. So, we know that there must be an $a$, where $a$ is odd. Now, we know that $F(a) - a$ will be even, as an even integer minus an even integer is even. So, the product listed above is even.

From here, I would assume that the result is true if $n$ is even, but I'm not 100% sure how to show it. I think it is a very similar process to the first part with taking out the $+1$ in the $n$ and cardinality of our set, as well as switching the words odd for even. Is this right?