A lot of proofs I have seen involve writing $p \in A_n$ as a product of an even number of transpositions. I have a different proof that involves disjointedness like in my question earlier today Prove $(\rho_1 \rho_2 \cdots \rho_r)^u = e \implies e=\rho_1^u=\rho_2^u=\cdots=\rho_r^u$.
Let $p \in S_n$.
If $p$ is a cycle, then $sign(p)=1 \iff p \in A_n \iff length(p)$ is odd, which can be seen from computing $\det(P)$, where $P$ is the associated permutation matrix to $p$. Write $p = (a_1 ... a_{2m+1}) = (a_1 a_2 a_3)(a_3 ... a_{2m+1})$ where $(a_3 ... a_{2m+1})$ has odd length because $(2m+1)-(3)+1=2m-1$, which is odd. Eventually, we can write every cycle of odd length as a product of 3-cycles.
If $p$ is not a cycle, then let its disjoint cycle decomposition be $p=\rho_1 \rho_2 ... \rho_r$.
If they are all cycles of odd length, then we are done.
If one has a cycle of even length, call it $\rho_{s_1}$. If $\rho_{s_1}$ is a 2-cycle, then by disjointedness, no other $\rho_{s_i}$ returns the switched indices, even though there is some other cycle of even length, call it $\rho_{s_1}$ (because $\det(\prod_{i=1}^m P_i) = \prod_{i=1}^{m}\det(P_i) = \prod_{i=1}^{m}(1) = 1$ implies that negative $\det(P_i)$'s come in pairs) and therefore, $p$ is somehow odd, a contradiction.
I also haven't worked out what to do if $\rho_{s_1}$ is an odd cycle of length 4,6,8, etc.
Any suggestions on how to continue?
Note: No using $(a b)(c d)=(a d c)(a b c)$ and $(a b)(b d) = (a b d)$, though I'm starting now to see why all the proofs I've found involve this.