Question: Can anyone prove that a [Weyl Algebra](http://en.wikipedia.org/wiki/Weyl_algebra) is not isomorphic to a matrix ring over a division ring?

Answer: 
Notation: The Weyl algebra is 
$$k[x\_1, x\_2, \ldots, x\_n, \partial\_1, \partial\_2, \ldots, \partial\_n]$$
with the obvious relations.

The Weyl algebra doesn't contain any division rings larger than $k$, and it is infinite dimensional over $k$. So, assuming you don't allow infinite matrices, that's a proof.

How to see that it doesn't contain any division ring larger than $k$? I just need to show that any nonconstant differential operator is not invertible. One way to see this is to notice that multiplying differential operators multiplies symbols, and the symbol of a nonconstant differential operator is a nonconstant polynomial.
A different proof would be to show that a Weyl algebra is not semisimple, that is, that it is not a direct sum of simple submodules as a left module over itself. However, note that there is an infinite descending chain of left submodules of a Weyl algebra given by 
$A\_n\supseteq A\_nd\supseteq A\_nd^2\supseteq A\_nd^3\supseteq...$
where $d$ is any non-invertible element. A direct sum of a finite number of simple modules can't have an infinite descending chain of submodules. Then, by the converse of Artin-Wedderburn, $A\_n$ is not a direct sum of matrix algebras over a divsion ring.

Of course, showing this sequence of submodules never stabilizes can be done by looking at the associated graded algebra, and noting that the $\overline{A\_nd^n}$ are always distinct there. However, then this answer starts getting closer to David's answer, so maybe this wasn't a truly different proof.