Let $S(n)$ be the symmetric group on $\{1,2,\ldots,n\}$. Let $\newcommand{\ss}{\sigma}\ss\in S(p+q)$ be the element with $$\ss(1)=p+1,\ss(2)=p+2,\ldots,\ss(q)=p+q,\ss(q+1)=1,\ss(q+2)=2,\ldots,\ss(p+q)=p.$$ I need to show that $\newcommand{\sn}{\operatorname{sign}}\sn\ss=(-1)^{pq}.$
Suppose $p=q$. Then $$\ss(1)=p+1,\ss(2)=p+2,\ldots,\ss(p)=p+p,\ss(p+1)=1,\ss(p+2)=2,\ldots,\ss(p+p)=p.$$ So $\ss=(1,p+1)(2,p+2)\cdots(p,p+p)$ and hence $\sn\ss=(-1)^p=(-1)^{p^2}$(because $p\equiv p^2{\mod 2}$).
Now WLOG, assume that $p\gt q$. So we can write $p=q+k$ for some $k\gt 0$. I aimed to write $\ss$ as a product of cycles. I have $\ss(p+1)=\ss(q+k+1)=k+1$, but I couldn't find $\ss(k+1)$. Any help is appreciated. Thank you.
The sign of a permutation $\sigma$ is also $(-1)^m$ where $m$ is the number of inversions in $\sigma$. (Wikipedia)
For the permutation in the question, the inversions are $(i,q+j)$ for $i=1,\dots,q$ and $j=1,\dots,p$ and so there are $pq$ inversions.
$$ \begin{pmatrix} 1&2&\cdots&q&q+1&q+2&\cdots&p+q \\ p+1&p+2&\cdots&p+q&1&2&\cdots&p\end{pmatrix} $$