Is there a slick proof of the classification of finitely generated abelian groups?
The slickest (nonconstructive!) proof I know of is the one I put in my Group Theory notes, p. 25. You choose a generating set $x_1,\ldots,x_n$ for the group such that $x_1$ has the minimum possible order, and then prove that the group is the direct sum of the subgroups generated by $x_1$ and by $x_2,\ldots,x_n$. Now apply induction on $n$ to see that the group is a direct sum of cyclic groups.
I reject the premise of the question. :-)
It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.
The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next.
Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.
Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )
The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.
The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □
Compare the complexity of this argument to the other arguments supplied so far.
Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.
Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.
The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.
Perhaps the issue is that the classification is not canonical or functorial in any reasonable way, and so any proof of this form must at some point create an arbitrary choice. (In particular, even though the classification tells us that every finite abelian group is isomorphic to its Pontryagin dual, there is no way to make this isomorphism canonical.) Presumably there is some category-theoretic way to formalise this issue, though I don't know how to do this. (A related fact, though, is that the above classification breaks down horribly for infinite abelian groups, much as the Jordan canonical form breaks down for infinite dimensional spaces.)
On the other hand, the special case of the classification for vector spaces over a finite field has the same issue (no canonical choice of basis), and yet doesn't seem to cause the same amount of dissatisfaction. I guess because here the full complexity of the Jordan canonical form does not emerge.
Greg Kuperberg pointed out to me in this blog post of mine that the Jordan canonical form for nilpotent transformations and the classification of abelian p-groups had essentially the same proof - in both cases the key is to understand the dynamics of a nilpotent homomorphism, which in the latter case is the operation of multiplication by p. This is perhaps the only "ugly" part of the whole story (and requires one to manually split a number of short exact sequences, etc.); reducing the Jordan normal form to the nilpotent case, or the classification of general abelian groups to p-groups, is all very clean and canonical.