I scribbled up some kind of proof some minutes ago.. I like it because it doesnt bring up
so many unnatural subjects.

If one has a linear homogeneous ode with characteristic equation (r-i)^2, then its solutions are (C+Dx)e^ix
The equation can be written u''-2iu'+u=0, where ' denotes differentiation with respect to x.
Then we just happen to try if (i sin(x) + cos(x)) satisfies the equation.. and it does.

Furthermore: all particular solutions to the equations can be written (C+Dx)e^ix,
so isinx+cosx = (C+Dx)e^ix, letting x=0 gives C=1, and by raising everything to 2 (don't know how to say that :P)

we get (isinx+cosx)^2=(cos^2x - sin^x) + 2icosxsinx = isin2x+cos2x = (1+2Dx)e^2ix = (1+2Dx+D^2 x^2)e^2ix, so obviously
D=0, and so we have isinx+cosx=e^ix

This proof (if it is a proof :P) uses complete solution to lin. hom. odes, common trig formulas, and a bit of simple algebra. I hope its valid.