A rotation symmetry in dimension 2 or 3 must move a lattice point to a succession of other lattice points in the same plane, generating a regular polygon of coplanar lattice points. We now confine our attention to the plane in which the symmetry acts , illustrated with lattice vectors in the figure.
Now consider an 8-fold rotation, and the displacement vectors between adjacent points of the polygon. If a displacement exists between any two lattice points, then that same displacement is repeated everywhere in the lattice. So collect all the edge displacements to begin at a single lattice point. The edge vectors become radial vectors, and their 8-fold symmetry implies a regular octagon of lattice points around the collection point. But this is impossible, because the new octagon is about 80% as large as the original. The significance of the shrinking is that it is unlimited. The same construction can be repeated with the new octagon, and again and again until the distance between lattice points is as small as we like; thus no discrete lattice can have 8-fold symmetry. The same argument applies to any k-fold rotation, for k greater than 6.
A shrinking argument also eliminates 5-fold symmetry. Consider a regular pentagon of lattice points. If it exists, then we can take every other edge displacement and (head-to-tail) assemble a 5-point star, with the last edge returning to the starting point. The vertices of such a star are again vertices of a regular pentagon with 5-fold symmetry, but about 60% smaller than the original.
Thus the theorem is proved.
The existence of quasicrystals and Penrose tilings shows that the assumption of a linear translation is necessary.  Penrose tilings may have 5-fold rotational symmetry and a discrete lattice, and any local neighborhood of the tiling is repeated infinitely many times, but there is no linear translation for the tiling as a whole.  And without the discrete lattice assumption, the above construction not only fails to reach a contradiction, but produces a (non-discrete) counterexample.  Thus 5-fold rotational symmetry cannot be eliminated by an argument missing either of those assumptions.  A Penrose tiling of the whole (infinite) plane can only have exact 5-fold rotational symmetry (of the whole tiling) about a single point, however, whereas the 4-fold and 6-fold lattices have infinitely many centres of rotational symmetry.