Prove that for each $\phi \in S_n$, $\det A=\sum_{\sigma \in S_n}\mathrm{sgn}(\sigma)\prod_{j=1}^n a_{\phi(j)\sigma(j)}$

Question

Assume that a permutation $\phi \in S_n$ is given.
Prove that :
$$\det A=\sum_{\sigma \in S_n}\operatorname{sgn} (\sigma)\space a_{\phi(1)\sigma(1)}a_{\phi(2)\sigma(2)} \cdots a_{\phi(n)\sigma(n)}$$

My Try :
The point is to choose an element from each row. So we have (am I correct?) :
$$\forall \phi \in S_n\space \quad \det A=\sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)\space a_{\phi(1)\sigma(\phi(1))}a_{\phi(2)\sigma(\phi(2))}\dots a_{\phi(n)\sigma(\phi(n))}$$

So, It holds for $\sigma \phi^{-1}$ as $\phi$.

Thus we have :
$$\det A=\sum_{\sigma\phi^{-1} \in S_n}\operatorname{sgn}(\sigma\phi^{-1})\space a_{\phi(1)\sigma\phi^{-1}(\phi(1))}a_{\phi(2)\sigma\phi^{-1}(\phi(2))}\dots a_{\phi(n)\sigma\phi^{-1}(\phi(n))}$$

$$=\sum_{\sigma\phi^{-1} \in S_n}\operatorname{sgn}(\sigma)\operatorname{sgn}\left(\phi^{-1}\right )\space a_{\phi(1)\sigma(1)}a_{\phi(2)\sigma(2)}\dots a_{\phi(n)\sigma(n)}$$

I don't know what to do next.

Answer 1

Note that $\{ (\phi(k), \sigma(k)) \}_k = \{ (k, \sigma(\phi^{-1}(k))) \}_k = \{ (k, (\sigma\circ \phi^{-1})(k)) \}_k$.

Then \begin{eqnarray} \sum_{\sigma \in S_n} \operatorname{sgn} (\sigma)\space a_{\phi(1)\sigma(1)}\cdots a_{\phi(n)\sigma(n)} &=& \sum_{\sigma \in S_n} \operatorname{sgn} (\sigma)\space a_{1 (\sigma\circ \phi^{-1})(1)}\cdots a_{n (\sigma\circ \phi^{-1})(n)} \\ &=& \sum_{\sigma \in S_n} \operatorname{sgn} (\sigma\circ \phi^{-1}) \operatorname{sgn} (\phi) \space a_{1 (\sigma\circ \phi^{-1})(1)}\cdots a_{n (\sigma\circ \phi^{-1})(n)} \\ &=& \operatorname{sgn} (\phi) \sum_{\sigma \in S_n} \operatorname{sgn} (\sigma\circ \phi^{-1}) \space a_{1 (\sigma\circ \phi^{-1})(1)}\cdots a_{n (\sigma\circ \phi^{-1})(n)} \\ &=& \operatorname{sgn} (\phi) \sum_{\sigma \in S_n} \operatorname{sgn} (\sigma)\space a_{1\sigma(1)}\cdots a_{n\sigma(n)}\\ &=& \operatorname{sgn} (\phi) \det A \end{eqnarray}