To see why they are both the same, first prove the following lemma:
Lemma 1
$$\mbox{sgn}(\sigma) =\prod_{1 \leq i < j \leq n} \frac{\sigma(j)-\sigma(i)}{ j-i}$$
To prove it, first note the the RHS must have absolute value 1, as every pair $(k,l)$ appears once at the bottom and once at the top. Therefore the RHS is $\pm 1$, and all you need to do is track the sign. Finally for the RHS you get a minus for each inversion.
Next, use this to get:
Lemma 2
$$\mbox{sgn}(\sigma \circ \tau) =\mbox{sgn}(\sigma)\mbox{sgn}( \tau)$$
Idea of proof: By Lemma 1:
$$
\mbox{sgn}(\sigma \circ \tau)=\prod_{1 \leq i < j \leq n} \frac{\sigma \circ \tau (j)-\sigma \circ \tau(i)}{ j-i} \\
=\prod_{1 \leq i < j \leq n} \frac{\sigma \circ \tau (j)-\sigma \circ \tau(i)}{ \tau (j)- \tau(i)} \cdot \prod_{1 \leq i < j \leq n} \frac{ \tau (j)- \tau(i)}{ j-i}=\mbox{sgn}(\sigma)\mbox{sgn}( \tau)
$$
by a similar computation to Lemma 1.
Lemma 3 If $\tau$ is a transposition $(i,j)$ then
$$\mbox{sgn}(\tau)=-1$$
Proof Just count the inversions.
Combining Lemma 3 and Lemma 2, you finally get:
Theorem Let $\sigma=\tau_1 \cdot ...\cdot \tau_k$ where $\tau_j$ are all transposition. Then
$$\mbox{sgn}(\sigma)=(-1)^k$$
Best Answer
Let $\pi=\langle p_1,p_2,\ldots,p_n\rangle$ be a permutation of $[n]=\{1,\ldots,n\}$. If $\pi$ has no inversions, it’s the identity permutation, which is even, so assume that $\pi$ has at least one inversion. Suppose that $1\le j<k\le n$, and $p_j>p_k$. Let $\rho=(p_j,p_k)\pi$ (where I apply the transposition first and then $\pi$); if $\rho=\langle r_1,\ldots,r_n\rangle$, then $r_j=p_k$, $r_k=p_j$, and $r_i=p_i$ if $i\in[n]\setminus\{j,k\}$. This swap obviously gets rid of the $\langle j,k\rangle$ inversion.
The only other inversions affected by this swap are those that involve position $j$ or $k$ and a position between $j$ and $k$. Suppose that $j<i<k$ and $\langle j,i\rangle$ is an inversion in $\pi$, i.e., $p_j>p_i$. Either $p_k>p_i$, in which case $\langle j,i\rangle$ is still an inversion in $\rho$, or $p_i>p_k$, in which case the swap gets rid of a $\langle j,i\rangle$ and an $\langle i,k\rangle$ inversion. Similarly, if the swap gets rid of an $\langle i,k\rangle$ inversion (instead of replacing it with a new one), it also gets rid of a $\langle j,i\rangle$ inversion. In all cases the number of inversions other than $\langle j,k\rangle$ that are affected by the swap is even. Thus, the total number of inversions affected by the swap is odd. Note also that the number of inversions is reduced by at least one.
Can you see how to finish it from here? I’ve left the conclusion of the argument in the spoiler-protected block below.