I think that the title already explains everything; I have been unable to prove that gray code is, for any dimension n for which n is a natural number, the only way to order all of the 2^n points in a 2^n grid so that consecutive points are separated only by one bit (including the starting and ending point). If anything requires clarification, I would be happy to do so.
It should be mentioned that the different iterations of Gray Code are identical in the context of my question as they are simply "rotations" of the n-dimensional structure which is created when linking together the points written out as 0's and 1's.
For example, if n=3, the corresponding Gray code may be 000 001 011 010 110 111 101 100 or 000 100 110 010 011 111 101 001, both of these are rotations of the same structure (namely the first iteration of the 3-dimensional Hilbert curve skullsinthestars.files.wordpress.com/2014/03/hilbertcurve3d.jpg) in three-dimensional space.
There are many Gray codes.
For three bits, there are $12$: you get a choice of which of three bits you wish to turn on first, then which of two bits you wish to turn on second, and finally whether you wish to turn on the third bit or turn off the first next.
These however are all isomorphic to each other: by permuting what order the bits are selected, or by choosing a different starting point, all of these cycles can be transformed into one another.
But this isn't true for four bits!
With four bits, there are $11$ distinct gray codes, listed below.
(0, 1, 3, 2, 6, 4, 5, 7, 15, 11, 9, 13, 12, 14, 10, 8) (0, 1, 3, 2, 6, 4, 5, 7, 15, 11, 10, 14, 12, 13, 9, 8) (0, 1, 3, 2, 6, 4, 5, 7, 15, 13, 12, 14, 10, 11, 9, 8) (0, 1, 3, 2, 6, 4, 5, 7, 15, 14, 10, 11, 9, 13, 12, 8) (0, 1, 3, 2, 6, 4, 5, 7, 15, 14, 12, 13, 9, 11, 10, 8) (0, 1, 3, 2, 6, 4, 12, 13, 5, 7, 15, 14, 10, 11, 9, 8) (0, 1, 3, 2, 6, 4, 12, 14, 10, 11, 15, 7, 5, 13, 9, 8) (0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 9, 11, 15, 14, 10, 8) (0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8) (0, 1, 3, 2, 6, 7, 5, 13, 9, 8, 10, 11, 15, 14, 12, 4) (0, 1, 3, 2, 6, 7, 5, 13, 15, 14, 10, 11, 9, 8, 12, 4)I'm uncertain how many there are in five bits; what I've seen suggests that the number reaches into the hundreds of thousands.