Here's an argument rather specific to $A_5$. Suppose you prove beforehand that $A_5$ is generated by its elements of order $p$ for each of $p=2,3,5$. Then a nontrivial proper quotient is also generated by its elements of order $p$ for each of $p=2,3,5$ and hence has order divisible by $2\times 3\times 5=30$. So we only have to discard normal subgroups of order 2, which amounts to the easy verification that there is no central element of order 2.

For the "beforehand" statement, the easiest is to check it's generated by 3-cycles (true in $A_n$ for all $n$), and then it is enough to write, in $A_5$, a single 3-cycle as product of 5-cycles, and as product of double transpositions.

(I didn't use it, but a side observation is that if a finite group $G$ is generated by its elements of order $p$ for every prime divisor of $|G|$, then every simple quotient of $G$ is nonabelian.)

Added (Oct 2019): as I said the argument is "rather specific" but still can be used in some other cases.

For the group of order $168=2^3.3.7$ a similar argument works. We also have to check that it is generated by its elements of order $p$ for each of $p=2,3,7$, which implies that every proper normal subgroup has index divisible by $2\times 3\times 7$, i.e. order dividing $4$. and then we have to discard a normal subgroup of order $2$ or $4$. Checking that the center is trivial is easy. The other case would be a non-central subgroup of order $4$, but then the centralizer of the latter would then have index $2$, $3$ or $6$, contradiction.

Next, in the case of $A_6$, the generation step is virtually already done in the $A_5$ case, since the primes are the same. Next a normal subgroup has index divisible by $30$, hence order dividing $12$. So, a minimal nontrivial normal subgroup has order $2$, $3$ or $4$ and again: show that the center is trivial, and if not central, the centralizer has too small index.

For $A_7$ one needs to play with elements of order $7$ in the generation step, and then every proper normal subgroup has order dividing $12$, which is ruled out exactly as in $A_6$.

For $A_8$ the argument shows its limits. While the generation step requires no further effort, the main issue will be to rule out specifically the existence of a nontrivial normal subgroup of order dividing $32$. Of course it can be done but the proof becomes quite inefficient.