Proving that for $\sigma\in S_n$ one has $\left|\prod_{iGood evening,
Could someone please demonstrate why this property is valid?

Given $\sigma\in S_n$
$$\left|\prod_{i<j} \frac{\sigma(j)-\sigma(i)}{j-i}\right|=1$$

 A: This is because $\sigma$ is a permutation, and therefore a one-to-one correspondence.  
You can rewrite the product in terms of numerators and denominators by way of
$$
\begin{align*}
\left\lvert\prod_{i<j}\frac{\sigma(j)-\sigma(i)}{j-i}\right\rvert&=\frac{\prod\limits_{i<j}\left\lvert\sigma(j)-\sigma(i)\right\rvert}{\prod\limits_{i<j}(j-i)}.
\end{align*}
$$
Re-index the product in the numerator by letting $h=\sigma^{-1}(i)$ and $k=\sigma^{-1}(j)$. Note that we can't assume $h<k$; however, we can still index the product in the numerator over all sets $\{h,k\}$ of two distinct integers in $[1,n]$. 
This reindexing yields
$$
\prod_{i<j}\lvert\sigma(j)-\sigma(i)\rvert=\prod_{\{h,k\}}\lvert k-h\rvert=\prod_{h<k}(k-h).
$$
A: Detailed proof: See Exercise 5.13 (a) in my Notes on the combinatorial fundamentals of algebra, 10th of January 2019. The claim I prove there is more general: I show that if $x_1, x_2, \ldots, x_n$ are any $n$ complex numbers, and if $\sigma$ is any permutation of $\left\{1,2,\ldots,n\right\}$, then
\begin{equation}
\prod_{i < j} \left(x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)}\right)
= \left(-1\right)^{\sigma} \cdot \prod_{i < j} \left(x_i - x_j\right) ,
\label{darij1.eq.1}
\tag{1}
\end{equation}
where $\left(-1\right)^{\sigma}$ denotes the sign of the permutation $\sigma$.
In order to obtain your equation from \eqref{darij1.eq.1}, you have to set $x_i = i$ and take absolute values (so that the sign $\left(-1\right)^{\sigma}$ disappears, since its absolute value is $1$).
Let me sketch how to quickly prove the weaker equality
\begin{equation}
\left|\prod_{i < j} \left(x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)}\right)\right|
= \left|\prod_{i < j} \left(x_i - x_j\right)\right|
\label{darij1.eq.2}
\tag{2}
\end{equation}
(which is still sufficient for your purposes). This is what @NickPeterson has already suggested, but my more rigorous notations shall hopefully close the cracks which let confusion slip through.
First of all, the absolute value of a product equals the product of the absolute values of the factors; thus,
\begin{align}
\left|\prod_{i < j} \left(x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)}\right)\right|
= \prod_{i < j} \left| x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)} \right| .
\end{align}
Next, let $P$ be the set of all pairs $\left(i, j\right)$ of integers $i, j \in \left\{1,2,\ldots,n\right\}$ satisfying $i < j$; also, let $G$ be the set of all $2$-element subsets of $\left\{1,2,\ldots,n\right\}$. Note that the product sign "$\prod\limits_{i < j}$" is equivalent to "$\prod\limits_{\left(i, j\right) \in P}$".
The two sets $P$ and $G$ have the same size (namely, $\dbinom{n}{2} = n\left(n-1\right) / 2$), and this is no coincidence: There is a bijection from $P$ to $G$. This bijection simply maps each pair $\left(i, j\right)$ to the two-element set $\left\{i, j\right\}$. The inverse of this bijection maps each two-element set to the pair consisting of its smaller element and its larger element (in this order).
The permutation $\sigma$ of $\left\{1,2,\ldots,n\right\}$ gives rise to a permutation $\sigma_*$ of the set $G$, which sends each two-element subset $I$ to $\sigma\left(I\right)$ (in other words, it sends each two-element subset $\left\{i,j\right\}$ to $\left\{\sigma\left(i\right), \sigma\left(j\right)\right\}$). Why is this a permutation of $G$? Well, again, its inverse is easy to find (it does the same thing, just with $\sigma^{-1}$ instead of $\sigma$). So $\sigma_*$ is a permutation of $G$, i.e., a bijection from $G$ to $G$.
Now the crucial insight: If $\left(i, j\right) \in P$, then the absolute value $\left| x_i - x_j \right|$ depends only on the set $\left\{i, j\right\} \in G$ (not on the pair $\left(i, j\right) \in P$). In other words, if $I \in G$ is any two-element subset, then we can define a real number $f_I$ by setting
\begin{align}
f_I = \left| x_i - x_j \right|,
\qquad \text{ where $I$ is written as $I = \left\{i, j\right\}$}.
\end{align}
In order to formally prove this, you should recall that there are exactly two ways of writing $I$ as $I = \left\{i, j\right\}$, and check that these two ways lead to the same value of $\left| x_i - x_j \right|$ (easy: these two ways only differ in the order of elements, and we have $\left| x_a - x_b \right| = \left| x_b - x_a \right|$).
Note that every $\left(i, j\right) \in P$ satisfies $\sigma_* \left( \left\{ i, j \right\} \right) = \left\{ \sigma\left(i\right), \sigma\left(j\right) \right\}$ and thus
\begin{align}
f_{\sigma_* \left( \left\{ i, j \right\} \right)}
= f_{\left\{ \sigma\left(i\right), \sigma\left(j\right) \right\}}
= \left| x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)} \right|
\label{darij1.eq.3}
\tag{3}
\end{align}
(by the definition of $f_{\left\{ \sigma\left(i\right), \sigma\left(j\right) \right\}}$).
Now,
\begin{align}
\left|\prod_{i < j} \left(x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)}\right)\right|
& = \prod_{i < j} \left| x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)} \right| \\
& = \prod_{\left(i, j\right) \in P} \underbrace{\left| x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)} \right|}_{\substack{ = f_{\sigma_* \left( \left\{ i, j \right\} \right)} \\ \left(\text{by \eqref{darij1.eq.3}}\right)}} \\
& \qquad \left(\text{since "$\prod\limits_{i < j}$" is equivalent to "$\prod\limits_{\left(i, j\right) \in P}$"}\right) \\
& = \prod_{\left(i, j\right) \in P} f_{\sigma_* \left( \left\{ i, j \right\} \right)} \\
& = \prod_{I \in G} f_{\sigma_* \left(I\right)}
\end{align}
(here, we have substituted $I$ for $\left\{ i, j \right\}$ in the product, since the map $G \to P, \  \left(i, j\right) \mapsto \left\{ i, j \right\}$ is a bijection).
Thus,
\begin{align}
\left|\prod_{i < j} \left(x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)}\right)\right|
= \prod_{I \in G} f_{\sigma_* \left(I\right)} = \prod_{I \in G} f_I
\label{darij1.eq.4}
\tag{4}
\end{align}
(here, we have substituted $I$ for $\sigma_* \left(I\right)$ in the product, since $\sigma_*$ is a bijection).
Note that the right hand side of \eqref{darij1.eq.4} does not depend on $\sigma$. Applying the same reasoning to the permutation $\operatorname{id}$ instead of $\sigma$, we thus obtain
\begin{align}
\left|\prod_{i < j} \left(x_{\operatorname{id}\left(i\right)} - x_{\operatorname{id}\left(j\right)}\right)\right|
= \prod_{I \in G} f_I .
\end{align}
In other words,
\begin{align}
\left|\prod_{i < j} \left(x_i - x_j\right)\right|
= \prod_{I \in G} f_I .
\end{align}
Comparing this equality with \eqref{darij1.eq.4}, we obtain
$\left|\prod_{i < j} \left(x_{\sigma\left(i\right)} - x_{\sigma\left(j\right)}\right)\right|
= \left|\prod_{i < j} \left(x_i - x_j\right)\right|$.
Thus, \eqref{darij1.eq.2} is proven.
