Just out of curiosity. Can the fact that the identity permutation is (only) even be proven by means of the sign function $\text{sgn}$?
In this and this post it was suggested that the simplest proof of the fact that the identity permutation can only be written as a product of an even number of transpositions is by means of $\text{sgn}$, although no proof was really specified. I've also seen proofs that use determinants such as this one.
The problem is that I do not see how one could define the sign function without first proving the identity is only an even permutation, similarly I do not know of a definition of determinants that do not appeal to this fact. I have tried, for example, to define $\text{sgn}$ as follows
Let $\rho \in \mathbb{S}_n$ and let $\sigma _1 \ldots \sigma_r$ be a product of transpositions with $r$ minimal. Then
$$\text{sgn}:=\begin{cases} 1 \ \ \ \ \text{ if } r \text{ is even} \\ -1 \ \text{ if } r \text{ is odd} \end{cases}$$
but then I seem to become unable to prove elementary properties of $\text{sgn}$ such as $\text{sgn}(\alpha \beta )=\text{sgn}(\alpha )\text{sgn}(\beta )$ which I believe are necessary to prove the main result.
Just associate each permutation $f$ with the result $r(f)$ of applying it to the terms of the finite sequence $(1,2,...,n)$, and define the sign of $f$ as the parity of the number of pairs $(x,y)$ such that $x < y$ but $x$ is after $y$ in $r(f)$. Then obviously the sign of $f$ is well-defined. And it is easy to prove that performing any swap on $r(f)$ changes the number of such pairs by an odd number, and hence the composition of $f$ with any swap has sign opposite to $f$. Done.