How may we formally prove the equivalence of taking permutations along the rows and along the columns in the Leibniz determinant? If the square matrix $\mathbf{A}\in M_{n\times n}$ and $a_{i,j}$ is the element in the $i$th row and the $j$th column of $\mathbf{A}$, and the sign of a permutation is positive when the number of interchanges is even, and negative when it is odd, and $\sigma,\tau$ are permutations in the permutation-group $S_n$ then:
$$\DeclareMathOperator{\sgn}{sgn}|\mathbf{A}|=\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{i=1}^na_{i,\sigma(i)}\equiv\sum_{\tau\in S_n}\sgn(\tau)\prod_{i=1}^na_{\tau(i),i}$$
And while it is intuitive to me that selecting elements row-wise or column-wise in the product comes to the same thing as everything is selected eventually, I can't see a clear proof for why the two expressions are equivalent, especially taking the sign, or "signature", of the permutation into account. Everything is selected, yet I don't see how to guarantee that all selections land in the same n-tuples, with the same sign! For example, a tuple of a $3\times3$ matrix might be $(a_{1,3},a_{2,1},a_{3,2})$, and using the other permutation method its correspondent is $(a_{2,1},a_{3,2},a_{1,3})$, but although these correspondents always exist, I cannot prove that the corresponding tuples have the same sign. Having the same sign would require an even number of interchanges between correspondents, which is clear in my example $((3,1,2)\to(2,3,1)$ with two interchanges$)$, but not so clear in the general case.
How can we show that the set of all the products $\sgn(\sigma)\cdot a_{i,\sigma(i)}$ is identical, (just with a different order), to the set of the products $\sgn(\tau)\cdot a_{\tau(i),i}$?

An argumentation is based upon the group structure of $S_n$. We start with
\begin{align*}
|\mathbf{A}|=\sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}\tag{1}
\end{align*}
Since the symmetric group $S_n$ with respect to composition of permutations is a group we have for each permutation $\sigma\in S_n$  a unique inverse $\sigma^{-1}\in S_n$.
If $\sigma(i)=k$ we have $\sigma^{-1}(k)=i$, and we get
\begin{align*}
a_{i,\sigma(i)}=a_{\sigma^{-1}(k),k}
\end{align*}
Since $\sigma\in S_n$ is a permutation of $\{1,\ldots,n\}$ and each element of $1,\ldots,n$ occurs precisely once we can write
\begin{align*}
\prod_{i=1}^n a_{\sigma^{-1}(i),i}\tag{2}
\end{align*}

Summing over all $n!$ permutations in $S_n$ it follows from (1) and (2):
\begin{align*}
\color{blue}{|\mathbf{A}|}&=\sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}\\
&=\sum_{\sigma\in S_n}\mathrm{sgn}(\sigma)\prod_{i=1}^na_{\sigma^{-1}(i),i}\\
&\,\,\color{blue}{=\sum_{\sigma^{-1}\in S_n}\mathrm{sgn}(\sigma^{-1})\prod_{i=1}^na_{\sigma^{-1}(i),i}}\tag{3}\\
\end{align*}
and the claim follows.

In (3) we use that $\mathrm{sgn}(\sigma)=\mathrm{sgn}(\sigma^{-1})$ and summing over $\sigma^{-1}\in S_n$ is just a reordering of summing over $\sigma \in S_n$.
