I asked a colleague who works in complex analysis and he also does not know any "integration free" proof or argument. However, he pointed me to a version free of complex numbers but equivalently astonishing (at least for me): [Weyl's Lemma][1], stating that any function  $u\in L^1_\text{loc}$ satisfying
$$
\int u\Delta \phi = 0
$$
for any test function $\phi$ is $C^\infty$. However, the proof also uses integration in the form of convolutions and hence, integrals are not at all avoided.

Another comment: Many theorems about (unsespected) smoothness of solutions of partial differential equation use some integral formula to deduce higher smoothness. One exception is the [Cauchy–Kowalevski theorem][2] but I don't see how this is related here.


  [1]: http://en.wikipedia.org/wiki/Weyl%27s_lemma_%28Laplace_equation%29
  [2]: http://en.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem