Simplicity of $A_n$
The one I like most goes roughly as follows (my reference is in French [Daniel Perrin, Cours d'algèbre], but maybe it's the one in Jacobson's Basic Algebra I) :
you prove that A(5) is simple by considering the cardinal of the conjugacy classes and seeing that nothing can be a nontrivial normal subgroup (because no nontrivial union of conjugacy classes including {id} has a cardinal dividing 60). Actually, you don't have to know precisely the conjugacy classes in A(5).
Then, you consider a normal subgroup N in A(n), n > 5 which is strictly larger than {id} and you prove (*) that it contains an element fixing at least n-5 points. The fact that A(5) is simple then gives that N contains every even permutation fixing the same n - 5 points. In particular, it contains a 3-cycle, and therefore contains all of A(n).
To prove (*), you consider a commutator [x,y], where x is nontrivial in your normal subgroup and y is a 3-cycle: by the very definition, it is the product of the 3-cycle and the conjugate of its inverse. So it's the product of two 3-cycles and has at the very least n-6 fixed points. But it's easy to see that you can chose the 3-cycle so that the commutator has n-5 fixed points (it is enough that the two 3-cycles have overlapping supports).
I like this proof because it keeps the "magical computation" part to a minimum, that simply amounts to the fact that you have automatically knowledge about a commutator if you have knowledge about one of his factors.
Here are the ingredients I think about:
$A_n$ is $n-2$-transitive on $\{1,2,\ldots,n\}$. That is, if $1\leq i_1\lt i_2\lt\cdots \lt i_{n-2}\leq n$ are any $n-2$ integers, and $j_1,\ldots,j_{n-2}$ are any $n-2$ distinct integers between $1$ and $n$, then there is an element of $A_n$ that maps $i_k$ to $j_k$ (in fact, one and only one element of $A_n$ that achieves this). This is easy: just write out the corresponding permutation. If it is even, this gives you an element of $A_n$. If it is not even, then adding a transposition involving the elements that do not occur among the $j_k$ makes it even and does not change the image of the $i_k$.
In particular, if $\sigma\in A_n$ fixes at least two elements, then every element with the same cycle structure to $\sigma$ is conjugate to $\sigma$ in $A_n$.
If $N$ is a normal subgroup of $A_n$, and $\sigma\in N$ fixes at least two elements, then every element with the same cycle structure as $\sigma$ is in $N$.
In particular, if $N$ is normal, contains a $3$-cycle, and $n\geq 5$, then $N$ contains all $3$-cycles; and since the $3$-cycles generate $A_n$ when $n\geq 3$, then $N=A_n$.
If $N$ is nontrivial and normal in $A_n$, and $n\geq 5$, then it contains a $3$-cycle: this involves a bit of manipulation, but the fact that $n\geq 5$ is as essential here as in (4) above: it gives you enough room to maneuver, room that you do not have in $A_4$ ($A_3$ and $A_2$ are also simple, but for silly reasons).
But I'm not sure if this particular line qualifies as "illuminating" in terms of giving you great insight into the structure of $A_n$; I feel they give me a good feel for what you can and cannot do in terms of playing with permutations (especially the transitivity, and the actual mechanics of point 5 above), and they remind me why $n\geq 5$ is important.