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 $\pi = \begin{pmatrix} 1 & 2 \end{pmatrix}$ be a transposition.

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

Then $\forall n \in \N_{>0}: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the factor $\paren {x_1 - x_2}$.

Thus:
 * $\begin{pmatrix} 1 & 2 \end{pmatrix} \cdot \Delta_n = - \Delta_n$

and thus $\begin{pmatrix} 1 & 2 \end{pmatrix}$ is odd.

From Conjugates of Transpositions:
 * $\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}$

Thus as $\begin{pmatrix} 2 & k \end{pmatrix}$ is self-inverse:
 * $\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1}$

But from Parity of Conjugate of Permutation:
 * $\map \sgn {\begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$

Thus:
 * $\sgn {\begin{pmatrix} 1 & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$

and thus $\begin{pmatrix} 1 & k \end{pmatrix}$ is odd.

Finally:

But from Parity of Conjugate of Permutation:
 * $\map \sgn {\begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$

Thus:
 * $\sgn {\begin{pmatrix} h & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$

and thus $\begin{pmatrix} h & k \end{pmatrix}$ is odd.