And how would one deduce this?
$[S_n, S_n]$ consists of even permutations so it's obvious that $[S_n, S_n] \leq A_n$, but is $[S_n, S_n] = A_n$ true as well? If so, how to deduce this? If not, how do we show that it's true for particular cases like $S_6$? I can only show that it's a subgroup.
In order to show that $A_n ≤ [S_n,S_n]$, you just have to write every $3$-cycle as a commutator in $S_n$, because the $3$-cycles generate $A_n$, if $n \geq 3$ (if $n \leq 2$, then this is trivial).
This is true because $(a \; b \; c) = (a\; c\; b)^2=(a\; c)(c\; b)(a\; c)(c\; b)$.