While trying to solve a bigger problem I came across this problem.
Let $\sigma \in S_n$ be defined as
$$\sigma = \begin{pmatrix} 1 & \cdots & p & p + 1 & \cdots & n \\ i_1 & \cdots & i_p & j_1 & \cdots & j_{n - p} \end{pmatrix}$$
Let $\tau$ be defined as
$$\tau = \begin{pmatrix} 1 & \cdots & n - p & n - p + 1 & \cdots & n \\ j_1 & \cdots & j_{n - p} & i_1 & \cdots & i_p \end{pmatrix}$$
We want to show that sgn($\sigma \tau) = (-1)^{p(n - p)}$. I thought that since for any $\gamma \in S_n$, sgn$(\gamma) = (-1)^{\text{#transpositions}}$, I could give an algorithm to convert $\sigma$ to $\tau$ using $p(n - p)$ transpositions, so that then we would have sgn$(\tau) =$ sgn$(\sigma) \cdot (-1)^{p(n - p)} \implies$ sgn$(\sigma \tau) = (-1)^{p(n - p)}$.
Here is the algorithm I came up with, shown for n = 5, p = 3 for the sake of simplicity:
$$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ i_1 & i_2 & j_1 & j_2 & j_3 \end{pmatrix}$$
$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ j_1 & i_2 & i_1 & j_2 & j_3 \end{pmatrix}$$
$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ j_1 & i_2 & j_2 & i_1 & j_3 \end{pmatrix}$$
$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ j_1 & i_2 & j_2 & j_3 & i_1 \end{pmatrix}$$
$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ j_1 & j_2 & i_2 & j_3 & i_1 \end{pmatrix}$$
$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ j_1 & j_2 & j_3 & i_2 & i_1 \end{pmatrix}$$
$$\tau = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ j_1 & j_2 & j_3 & i_1 & i_2 \end{pmatrix}$$
As you can see, it is easy to understand. However it seems hard to formalize without getting lost in an ocean of notation (and even harder to understand if formalized). For this reason I was wondering if anyone would be capable of providing a more elegant answer to this problem.
Note: You will have noticed that the title to the question does not manage to describe problem properly, but I did not come up with a better way of describing it. Any suggestions are welcome.
I'm not sure I understand the problem, but anyway, the sign of $\sigma\tau$ is the same as that of $\rho=\tau^{-1}\circ\sigma$. Now $\rho$ takes $1,2,\ldots,n$ to $n-p+1,n-p+2,\ldots,n,1,2,\ldots,n-p$. It really "cuts" a pack of cards, moving the last $p$ cards to the stop. Anyway, it's plain that $\rho$ has $p(n-p)$ inversions, so its sign is $(-1)^{p(n-p)}$.