[proofplan] We write the transposition as the swap of two indices and, after ordering them, call the smaller one $i$ and the larger one $j$. We then compute its inversion number directly. The only inversions are the pair $(i,j)$ itself and the pairs involving exactly one of the intermediate indices between $i$ and $j$. This gives $2(j-i)-1$ inversions, which is odd, so the defining formula for the signature gives $\operatorname{sgn}(\tau)=-1$. [/proofplan]

[step:Write the transposition as a swap of two ordered indices] Since $\tau\in S_n$ is a transposition, there exist distinct indices $i,j\in\{1,\dots,n\}$ such that $\tau$ swaps $i$ and $j$ and fixes every element of $\{1,\dots,n\}\setminus\{i,j\}$. Interchanging the labels if necessary, assume $i<j$. Thus \begin{align*} \tau(i)=j,\qquad \tau(j)=i, \end{align*} and, for every $r\in\{1,\dots,n\}\setminus\{i,j\}$, \begin{align*} \tau(r)=r. \end{align*} [/step]

[step:Count exactly the inversions of the transposition]Let $I(\tau)$ denote the inversion set of $\tau$, namely \begin{align*} I(\tau):=\{(a,b)\in\{1,\dots,n\}^2:a<b\text{ and }\tau(a)>\tau(b)\}. \end{align*} We claim that \begin{align*} I(\tau)=\{(i,j)\}\cup\{(i,k):i<k<j\}\cup\{(k,j):i<k<j\}. \end{align*} Indeed, the pair $(i,j)$ is an inversion because $\tau(i)=j>i=\tau(j)$. If $i<k<j$, then $(i,k)$ is an inversion because $\tau(i)=j>k=\tau(k)$, and $(k,j)$ is an inversion because $\tau(k)=k>i=\tau(j)$. Conversely, let $(a,b)\in I(\tau)$. If neither $a$ nor $b$ belongs to $\{i,j\}$, then $\tau(a)=a<b=\tau(b)$, contradicting that $(a,b)$ is an inversion. If $b=i$, then $a<i$ and $\tau(a)=a<j=\tau(i)$, so $(a,i)$ is not an inversion. If $a=j$, then $b>j$ and $\tau(j)=i<b=\tau(b)$, so $(j,b)$ is not an inversion. If $a=i$ and $b>j$, then $\tau(i)=j<b=\tau(b)$, so $(i,b)$ is not an inversion. If $a<i$ and $b=j$, then $\tau(a)=a<i=\tau(j)$, so $(a,j)$ is not an inversion. The only remaining possibilities are $(i,j)$, $(i,k)$ with $i<k<j$, and $(k,j)$ with $i<k<j$, which are precisely the pairs already listed. Therefore \begin{align*} |I(\tau)|=1+(j-i-1)+(j-i-1)=2(j-i)-1. \end{align*}[/step]

[guided]We compute the inversion set rather than using an abstract parity argument, because the transposition has only two moved points and its inversions can be listed explicitly. Define \begin{align*} I(\tau):=\{(a,b)\in\{1,\dots,n\}^2:a<b\text{ and }\tau(a)>\tau(b)\}. \end{align*} Since $\tau$ swaps $i$ and $j$ with $i<j$, we have $\tau(i)=j$, $\tau(j)=i$, and $\tau(r)=r$ for all $r\notin\{i,j\}$. First, we identify the inversions that definitely occur. The pair $(i,j)$ is an inversion because \begin{align*} \tau(i)=j>i=\tau(j). \end{align*} Next take an intermediate index $k\in\{1,\dots,n\}$ satisfying $i<k<j$. Since $k$ is fixed by $\tau$, the pair $(i,k)$ is an inversion: \begin{align*} \tau(i)=j>k=\tau(k). \end{align*} The pair $(k,j)$ is also an inversion: \begin{align*} \tau(k)=k>i=\tau(j). \end{align*} Thus every pair in \begin{align*} \{(i,j)\}\cup\{(i,k):i<k<j\}\cup\{(k,j):i<k<j\} \end{align*} belongs to $I(\tau)$. Now we rule out all other pairs. Let $(a,b)$ be a pair with $a<b$. If neither $a$ nor $b$ is one of the swapped indices, then both are fixed, so \begin{align*} \tau(a)=a<b=\tau(b), \end{align*} and no inversion occurs. If $b=i$, then $a<i$, and \begin{align*} \tau(a)=a<j=\tau(i), \end{align*} so $(a,i)$ is not an inversion. If $a=j$, then $b>j$, and \begin{align*} \tau(j)=i<b=\tau(b), \end{align*} so $(j,b)$ is not an inversion. If $a=i$ and $b>j$, then \begin{align*} \tau(i)=j<b=\tau(b), \end{align*} so $(i,b)$ is not an inversion. If $a<i$ and $b=j$, then \begin{align*} \tau(a)=a<i=\tau(j), \end{align*} so $(a,j)$ is not an inversion. The only remaining possible inversions are therefore $(i,j)$, the pairs $(i,k)$ with $i<k<j$, and the pairs $(k,j)$ with $i<k<j$. There are $j-i-1$ intermediate indices $k$ satisfying $i<k<j$, so \begin{align*} |I(\tau)|=1+(j-i-1)+(j-i-1)=2(j-i)-1. \end{align*}[/guided]

[step:Apply the definition of signature to the odd inversion count] By the definition of the signature in terms of inversions, \begin{align*} \operatorname{sgn}(\tau)=(-1)^{|I(\tau)|}. \end{align*} Using the computed value $|I(\tau)|=2(j-i)-1$, we obtain \begin{align*} \operatorname{sgn}(\tau)=(-1)^{2(j-i)-1}. \end{align*} Since $2(j-i)-1$ is odd, this gives \begin{align*} \operatorname{sgn}(\tau)=-1. \end{align*} This proves that every transposition in $S_n$ has negative signature. [/step]