Alternative proofs that A5A_5 is simple

I'm very happy to show this proof, which makes use of a technique frequently employed in a recent paper of mine. It's my favorite proof that $A_5$ is not solvable, which as you pointed out in your last bullet proves that $A_5$ is simple.

Definition. For $n\in \mathbb{N}$, denote by $\pi(n)$ the set of prime divisors of $n$. The prime graph of a finite group $G$, denoted $\Gamma_G$, is the graph with vertex set $\pi(|G|)$ with an edge between primes $p$ and $q$ if and only if there is an element of order $pq$ in $G$.

By Lucido (1999), Prop. 1, the complement of the prime graph of a solvable group $G$ is triangle-free. It is obvious from cycle types that $A_5$ contains no elements of order $6,10,$ or $15$, so $\Gamma_{A_5}$ is the empty graph on three vertices. Therefore, $A_5$ is not solvable.