Transposition is of Odd Parity

Theorem
Let $S_n$ denote the set of permutations on $n$ letters.

Let $\pi \in S_n$ be a transposition.

Then $\pi$ is of odd parity.

Proof
Let $\pi = \begin{bmatrix} 1 & 2 \end{bmatrix}$ be a transposition.

Let $\Delta_n$ be defined as Product of Differences.

Then $\forall n \in \N^*: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the factor $\left({x_1 - x_2}\right)$.

Thus $\begin{bmatrix} 1 & 2 \end{bmatrix} \cdot \Delta_n = - \Delta_n$, and thus $\begin{bmatrix} 1 & 2 \end{bmatrix}$ is odd.


 * From Conjugates of Transpositions, $\begin{bmatrix} 1 & k \end{bmatrix} = \begin{bmatrix} 2 & k \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & k \end{bmatrix}$.

Thus as $\begin{bmatrix} 2 & k \end{bmatrix}$ is self-inverse, $\begin{bmatrix} 1 & k \end{bmatrix} = \begin{bmatrix} 2 & k \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & k \end{bmatrix}^{-1}$.

But from Parity of Conjugate of Permutation, $\operatorname{sgn} \left({\begin{bmatrix} 2 & k \end{bmatrix} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & k \end{bmatrix}^{-1}}\right) = \operatorname{sgn} \left({\begin{bmatrix} 1 & 2 \end{bmatrix}}\right)$.

Thus $\operatorname{sgn} \left({\begin{bmatrix} 1 & k \end{bmatrix}}\right) = \operatorname{sgn} \left({\begin{bmatrix} 1 & 2 \end{bmatrix}}\right)$ and thus $\begin{bmatrix} 1 & k \end{bmatrix}$ is odd.


 * Finally:

But from Parity of Conjugate of Permutation, $\operatorname{sgn} \left({\begin{bmatrix} 1 & h \end{bmatrix} \begin{bmatrix} 1 & k \end{bmatrix} \begin{bmatrix} 1 & h \end{bmatrix}^{-1}}\right) = \operatorname{sgn} \left({\begin{bmatrix} 1 & k \end{bmatrix}}\right)$.

Thus $\operatorname{sgn} \left({\begin{bmatrix} h & k \end{bmatrix}}\right) = \operatorname{sgn} \left({\begin{bmatrix} 1 & k \end{bmatrix}}\right)$ and thus $\begin{bmatrix} h & k \end{bmatrix}$ is odd.