Proofs of the Stallings-Swan theorem It is a well-known and deep${}^\ast$ theorem that if a group $G$ has cohomological dimension one then it must be free. This was proved in the late 60's by Stallings (for finitely generated groups) and Swan.
The proofs in the original articles are well-written and informative, bringing together a lot of ideas from topology, algebra and set theory. However there seems to be some scope for the proofs to be shortened, for instance by a clever choice of projective resolution, or the use of non-abelian cohomology.
My questions then is

Have any alternative (shorter) proofs of the Stallings-Swan theorem appeared since 1969?

${}^\ast$I am prepared to accept that the answer may be 'no', but in that case I wonder if someone could offer an explanation of why this theorem is so "deep", ie why there cannot exist some "quick trick" proof.
 A: The heart of the matter is the Stallings' "ends of groups" theorem: A finitely-generated group with infinitely many ends splits as graph of groups with finite edge groups. In addition, one also has to show that the decomposition process terminates for your group (this property is called accessibility). Neither one has a quick an dirty proof and for a good reason. 
a. Dunwoody has shown that accessibility fails for some finitely-generated groups with torsion, so something interesting (Grushko's theorem) is going on even in the easy part of the proof. 
b. The ends of groups deals with the key problem of geometric group theory: Relating geometric properties of a group and its algebraic structure. Each time one manages to recover algebraic structure of a group from geometric information about the group, some minor (or major) miracle has to happen and, to the best of my knowledge (with few trivial exceptions) there are no easy proofs of the results of this type. 
The "shortest" proof of the "Ends of groups theorem" is due to Gromov, see pages 228-230 of his essay on hyperbolic groups. The trouble with Gromov's proof is that it relies  on a compactness property  ("obvious" to Gromov) for a certain family of harmonic functions (in addition, a construction of the tree was missing in his proof, and this requires some trickery if one uses harmonic functions for finitely-generated groups). This compactness result (as far as I know) has no easy proof. I wrote a (somewhat long) proof in http://arxiv.org/pdf/0707.4231.pdf , Bruce Kleiner managed to shorten it to about 7 pages (this is not published), but his proof is still not quick. 
Dunwoody's proof (see John Klein's excellent comments) improved on the Stallings' proof, but his proof is still quite complicated. 
Niblo's proof in http://eprints.soton.ac.uk/29820/1/Stallingstheorem.pdf
provides another geometric argument using Sageev's complex, but Niblo's paper is still 20 pages long.    
Just to indicate how nontrivial Stallings' theorem is, consider the question: Is it true that every finitely generated group $G$ of homological dimension 1 is free? This is false for infinitely generated groups (like $G={\mathbb Q}$) and is true for finitely-presented groups (simply since in this case cohomological dimension is also 1). Otherwise, this problem is open since Stallings' theorem (and not for the lack of trying!).  
