Sign of Permutation is Plus or Minus Unity

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Let $\left \langle {x_k} \right \rangle_{k \in \N^*_n}$ be a finite sequence in $\R$.

Let $\pi \in S_n$.

Let $\Delta_n \left({x_1, x_2, \ldots, x_n}\right)$ be the product of differences of $\left({x_1, x_2, \ldots, x_n}\right)$.

Let $\operatorname{sgn} \left({\pi}\right)$ be the sign of $\pi$.

Let $\pi \cdot \Delta_n \left({x_1, x_2, \ldots, x_n}\right)$ be defined as:
 * $\pi \cdot \Delta_n \left({x_1, x_2, \ldots, x_n}\right) = \Delta_n \left({x_{\pi \left({1}\right)}, x_{\pi \left({2}\right)}, \ldots, x_{\pi \left({n}\right)}}\right)$

Then either:
 * $\pi \cdot \Delta_n = \Delta_n$

or:
 * $\pi \cdot \Delta_n = -\Delta_n$

That is:


 * $\operatorname{sgn} \left({\pi}\right) = \begin{cases}

1 & :\pi \cdot \Delta_n = \Delta_n \\ -1 & : \pi \cdot \Delta_n = -\Delta_n \end{cases}$

Thus:
 * $\pi \cdot \Delta_n = \operatorname{sgn} \left({\pi}\right) \Delta_n$

Proof
If $\exists i, j \in \N^*_{\le n}$ such that $x_i = x_j$, then $\Delta_n \left({x_1, x_2, \ldots, x_n}\right) = 0$ and the result follows trivially.

So, suppose all the elements $x_k$ are distinct.

Let us use $\Delta_n$ to denote $\Delta_n \left({x_1, x_2, \ldots, x_n}\right)$.

Let $1 \le a < b \le n$.

Then $x_a - x_b$ is a divisor of $\Delta_n$.

Then $x_{\pi \left({a}\right)} - x_{\pi \left({b}\right)}$ is a factor of $\pi \cdot \Delta_n$.

There are two possibilities for the ordering of $\pi \left({a}\right)$ and $\pi \left({b}\right)$:

Either $\pi \left({a}\right) < \pi \left({b}\right)$ or $\pi \left({a}\right) > \pi \left({b}\right)$.

If the former, then $x_{\pi \left({a}\right)} - x_{\pi \left({b}\right)}$ is a factor of $\Delta_n$.

If the latter, then $- \left({x_{\pi \left({a}\right)} - x_{\pi \left({b}\right)}}\right)$ is a factor of $\Delta_n$.

The same applies to all factors of $\Delta_n$.

Thus:


 * $\displaystyle \pi \cdot \Delta_n = \pi \cdot \prod_{1 \le i < j \le n} \left({x_i - x_j}\right) = \pm \prod_{1 \le i < j \le n} \left({x_i - x_j}\right) = \pm \Delta_n$