I have seen two proofs of the simplicity of $A_n,~ n \geq 5$ (Dummit & Foote, Hungerford). But, neither of them are such that they 'stick' to the head (at least my head). In a sense, I still do not have the feeling that I know why they are simple and why should it be 5 and not any other number (perhaps this is only because 3-cycles become conjugate in $A_n$ after $n$ becomes greater than 4).
What is the most illuminating proof of the simplicity of $A_n,~ n \geq 5$ that you know?
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.
The way to understand this theorem is to understand the following statement:
To illustrate what I mean, I will prove the following theorem, which is related to but slightly easier than the simplicity of $A_n$:
The main idea of the proof is to imagine what a normal subgroup of $S_n$ must look like:
Any normal subgroup of $S_n$ must be a union of conjugacy classes.
But each conjugacy class of $S_n$ is simply the set of all permutations with a given cycle structure.
Therefore, a normal subgroup of $S_n$ must consist of a collection of possible cycle structures which is closed under multiplication.
For example, the conjugacy classes in $S_4$ are the following:
Any normal subgroup in $S_4$ would have to be the union of some of these, and would have to be closed under multiplication. It's easy to check that the only possibilities are $A_4 = (1) \cup (3) \cup (5)$ and the 4-element subgroup $(1) \cup (5)$.
Similarly, the conjugacy classes in $S_5$ are the following:
It's easy to check that the only finite union of these that is closed under multiplication is $A_5$.
At this point, the rest of the proof just involves playing around with permutations and cycle structures. In particular, it's not too hard to prove the following statement:
Thus any normal subgroup of $S_n$ must either contain all of the 3-cycles, or all elements of the form $(\ast\;\ast)(\ast\;\ast)$. For $n \geq 5$, elements of these forms generate all of $A_n$, and therefore $A_n$ is the only nontrivial normal subgroup of $S_n$. (Since $A_n$ has index two, any nontrivial subgroup that contains $A_n$ must equal $A_n$.)
The proof that $A_n$ is simple is fairly similar, but complicated by the fact that the conjugacy classes in $A_n$ aren't precisely the cycle structures. (Some cycle structures in $A_n$ have two corresponding conjugacy classes.) Alternatively, it may be possible to prove that any normal subgroup of $A_n$ would also have to be normal in $S_n$, and then apply the above theorem to conclude that $A_n$ is simple.
Many proofs of this theorem involve a trick, which is not so natural to think, or remember (I feel).
But the following is a Lemma, which is very easy to state, easy to prove, and immediately implies simplicity of $A_n$. This I saw as an exercise in Wilson's Finite Simple Groups.
Lemma: Let $n\geq 5$. Then every non-trivial conjugacy class in $A_n$ contains at least $n$ elements.
Proof: Consider a non-identity $\sigma\in A_n$. Two cases arises: $\sigma$ contains at least a cycle of length $\geq 3$ or all cycles have length $2$.
Key: if $\sigma$ takes $i$ to $j$ (i.e. $\sigma(i)=j$) then $\tau\sigma\tau^{-1}$ takes $\tau(i)$ to $\tau(j)$: $$\tau\sigma\tau^{-1}(\tau(i))=\tau(j).$$
(A) $\sigma=(123\cdots)\cdots$. Let $\tau=(234)$. Then $$\tau\sigma\tau^{-1}=(134\cdots)\cdots$$ This is a conjugate of $\sigma$ and is clearly different from $\sigma$. Similarly, taking $\tau=(23k)$ with $k=4,5,\cdots,n$, we get $n-3$ conjugates of $\sigma$ different from $\sigma$ (i.e. we got $n-2$ conjugates). Then slightly change $\tau$'s as $\tau=(2k3)$, with $k=4,5$ (since $n$ is at least five), we get two more conjugates of $\sigma$. This case is complete.
(B) $\sigma=(12)(34)\cdots$ Again for $\tau=(234), (235), \cdots, (23n)$, we get $\tau\sigma\tau^{-1}$ equal to $$ (13)(42), (13)(54), \cdots (13)(n4);$$ these $n-3$ distinct conjugates of $\sigma$ together with $\sigma$ give $n-2$ elements in conjugacy class of $\sigma$. You may try to obtain two more, just by slight modification [as in Case A]. Q.E.D.
Proof of Theorem: We assume simplicity of $A_5$ is proved. Let $n>5$ and $N$ a normal subgroup of $A_n$. Then $N\cap A_{n-1}$ is normal in $A_{n-1}$ ($A_{n-1}$ is subgroup of permutations fixing $n$ in $A_n$). By induction, $N\cap A_{n-1}=A_{n-1}$ or $N\cap A_{n-1}=1$. In the first case, $A_{n-1}\subseteq N$, and hence $N$ contains a $3$-cycle. This implies that $N=A_n$. In the second case, i.e. $N\cap A_{n-1}=1$, we get $|N|\leq n$ (since $NA_{n-1}\leq A_n$ hence $\frac{|N|.|A_{n-1}|}{|N\cap A_{n-1}|}\leq |A_n|$).
Thus $N$ is normal subgroup of order $\leq n$, it follows that it should contain a conjugacy class of size $<n$ unless $N=1$. We arrive either at contradiction or conclusion that $N=1$. QED.
One may say that proof is lengthy; but I would not say. I would say, it is easier because, it is based on two facts which are easy to prove even after a long break from the topic.
If $\sigma(i)=j$ then $\tau\sigma\tau^{-1}(\tau(i))=\tau(j)$.
For $n\geq 5$, any non-trivial conjugacy class in $A_n$ contains at least $n$ elements.
The step-by-step verification of these proves the theorem.
Since LIE's answer reduces one to the case of $A_5$, I like identifying $A_5$ with the icosahedral group and using the geometry of the icosahedron to make the calculations of the number of elements conjugate to those leaving vertices, edges, and faces, respectively fixed. This is on page 20 of the notes for math 843-1 here.
To identify Icos with $A_5$, note the 30 edges are partitioned into 5 subsets of size 6, by looking at the 5 sets of orthogonal triples in the 15 lines joining the centers of pairs of opposite edges.
Then LIE's answer (and Arturo's) shows why $A_5$ is really the key case. Of course the apparently ad hoc geometry used here is mirrored abstractly by the conjugacy classes of elements in the group. I.e. every group acts on itself by conjugation, and sometimes we can represent that action by a geometric figure.
