\begin{align*} \det\left(\boldsymbol{a}_{\tau\left(1\right)},\boldsymbol{a}_{\tau\left(2\right)},\cdots,\boldsymbol{a}_{\tau\left(n\right)}\right) & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{1,\sigma\left(\tau\left(1\right)\right)}a_{1,\sigma\left(\tau\left(2\right)\right)}\cdots a_{n,\sigma\left(\tau\left(n\right)\right)}\\ & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{1,\left(\sigma t\right)\left(1\right)}a_{1,\left(\sigma t\right)\left(2\right)}\cdots a_{n,\left(\sigma t\right)\left(n\right)}\\ & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\tau\tau^{\bullet}\right)a_{1,\left(\sigma t\right)\left(1\right)}a_{1,\left(\sigma t\right)\left(2\right)}\cdots a_{n,\left(\sigma t\right)\left(n\right)}\\ & =\sgn\left(\tau^{\bullet}\right)\sum_{\sigma\in S_{n}}\sgn\left(\sigma\tau\right)a_{1,\left(\sigma t\right)\left(1\right)}a_{1,\left(\sigma t\right)\left(2\right)}\cdots a_{n,\left(\sigma t\right)\left(n\right)}\\ & =\sgn\left(\tau\right)\sum_{\sigma\tau\in S_{n}}\sgn\left(\sigma\tau\right)a_{1,\left(\sigma t\right)\left(1\right)}a_{1,\left(\sigma t\right)\left(2\right)}\cdots a_{n,\left(\sigma t\right)\left(n\right)}\\ & =\sgn\left(\tau\right)\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{n}\right) \end{align*} 途中で任意の\(\tau\in S_{n}\)に対し、\(\sigma\)が\(S_{n}\)全体を動くとき、\(\sigma\tau\)も\(S_{n}\)全体を動くことを使った。
何故なら\(\sigma_{1},\sigma_{2}\in S_{n}\)のとき、\(\sigma_{1}\tau=\sigma_{2}\tau\)ならば\(\sigma_{1}\tau\tau^{\bullet}=\sigma_{2}\tau\tau^{\bullet}\)より、\(\sigma_{1}=\sigma_{2}\)となり、また\(\sigma_{1}=\sigma_{2}\Rightarrow\sigma_{1}\tau=\sigma_{2}\tau\)が成り立つので、\(\sigma_{1}=\sigma_{2}\Leftrightarrow\sigma_{1}\tau=\sigma_{2}\tau\)となる。
これより、\(S_{n}\rightarrow S_{n}\tau,\sigma\mapsto\sigma\tau\)が単射、\(S_{n}\tau\rightarrow S_{n},\sigma\tau\mapsto\sigma\)が単射となり、\(S_{n}\rightarrow S_{n}\tau\)が全単射となるからである。

-

行列式を行で展開すると次のようになる。
\begin{align*} \det\left(\boldsymbol{a}_{\tau\left(1\right)},\boldsymbol{a}_{\tau\left(2\right)},\cdots,\boldsymbol{a}_{\tau\left(n\right)}\right) & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\tau\left(1\right),\sigma\left(1\right)}a_{\tau\left(2\right),\sigma\left(2\right)}\cdots a_{\tau\left(n\right),\sigma\left(n\right)}\\ & =\sum_{\sigma\in S_{n}}\sgn\left(\sigma\tau\right)a_{\tau\left(1\right),\sigma\tau\left(1\right)}a_{\tau\left(2\right),\sigma\tau\left(2\right)}\cdots a_{\tau\left(n\right),\sigma\tau\left(n\right)}\\ & =\sgn\left(\tau\right)\sum_{\sigma\in S_{n}}\sgn\left(\sigma\right)a_{\tau\left(1\right),\sigma\tau\left(1\right)}a_{\tau\left(2\right),\sigma\tau\left(2\right)}\cdots a_{\tau\left(n\right),\sigma\tau\left(n\right)}\\ & =\sgn\left(\tau\right)\det\left(\boldsymbol{a}_{1},\boldsymbol{a}_{2},\cdots,\boldsymbol{a}_{n}\right) \end{align*}