The proof in Rotman's book, Introduction to the Theory of Groups, that $A_n$ is simple consists of the observation that $A_n$ is generated by the $3$-cycles, and hence that if a normal subgroup $H\lhd A_n$ contains one, it is all of $A_n$, for all $3$-cycles are conjugate in $A_n$. One then proves $A_5$ is simple by counting, say, and that $A_6$ is simple by showing any nontrivial normal subgroup contains a $3$-cycle, as follows: if $1\neq\alpha \in H$ fixes a symbol, say $i$; let $F=\{\beta\in A_6:\beta i=i\}$. Then $F\simeq A_5$, $\alpha\in F$ and $\alpha\in F\cap H\lhd F$, so by simplicity $F\cap H=F$, this means $F\leqslant H$, and $H$ contains a $3$-cycle.
One then shows $H$ cannot consists only of elements with empty support by looking at $(12)(3456)$ and $(123)(456)$. In the first case, note $\alpha^2$ fixes $1$ and $2$, and in the second case if we let $\beta=(234)$, $[\alpha,\beta]$ is not $1$; and fixes $6$. So no $H$ exists.
Finally, one shows $A_n$ is simple by a similar token: let $H\lhd A_n$ nontrivial and pick $\beta\neq 1$; say $\beta i=j\neq i$. If $\alpha$ is a $3$-cycle that fixes $i$ and moves $j$, $[\alpha,\beta]\neq 1$ is in $H$, and is the product of two $3$ cycles. Thus it moves at most six symbols, and if we let $F$ be the set of even permutations that fix every other symbol, $F\simeq A_6$, and $[\alpha,\beta]\in H\cap F\lhd F$, so again since $A_6$ is simple, $F\leqslant H$ and $H$ contains a $3$-cycle.
I am curious about who this proof is due to. Some reference or literature would be nice.
Add $A_n$ is generated by the $3$-cycles because $(ij)(jk)=(ijk)$, and $(ij)(kl)=(ijk)(jkl)$ and every element of $A_n$ is an even product of permutations. Every $3$-cycle in $A_n$ is conjugate: one shows every $3$-cycle in $A_5$ is conjugate, and then shows $(123)$ and $(ijk)$ are conjugate in $A_n$ if they are not disjoint, for they move at most $5$ symbols, hence sit in $A_5$. Else, we note by the previous argument $(123)\sim (3jk)\sim (ijk)$.