Shapes Filling Space

As many wargamers know, you cannot fill up a flat surface with any regular polygon with more than six sides. Hexes work; squares work; triangles work. I'm not sure about pentagons, but septagons, octagons etc do NOT work.

Can this sort of thought be extended to three dimensions? What are the three dimensional versions of regular polygons? Are they the Platonic Solids? If so, why are there only five of the latter, but infinitely many of the former?

http://www.google.co.il/search?hl=iw&q=space-filling+polyhedron&meta=

http://en.wikipedia.org/wiki/Platonic_solid#Geometric_proof

in 4 dimensions there are 6 regular solids, in 5,6,7, etc. dimensions there are 5:

http://en.wikipedia.org/wiki/Convex_regular_4-polytope

A pentagon cannot tile a 2-dimensional plane because 360 degrees does not divide evenly into the interior angle of a regular pentagon.

As for 3-dimensional solids with identical regular polygons, there are only 5, and you can find them in the following manner.

For each of them, Each vertex looks identical. That is to say that each one has a specific number of a specific shape attached to it. The sum of the interiors angles has to be less than 360 degrees. For instance, no such solid has hexagonal sides, because int interios angles of 3 hexagons sums to exactly 360 degrees, while tiles the plane, but doesn't "curve" to form a rounded shape, and thus extends infinitely far.

If 3 triangles meet at each vertex, you get a tetrahedron (4 sides).
If 4 triangles meet at each vertex, you get a octohedron (8 sides).
If 5 triangles meet at each vertex, you get a dodecahedron (20 sides).
6 triangles tile the plane, and 7+ are impossible.
If 3 squares meet at each vertex, you get a cube.
4 squares tile the plane, and 5+ are impossible.
If 3 pentagons meet at each vertex, you get an icosahedron (12 sides).
4+ pentagons are impossible.
3 hexagons tile the plane and 4+ are impossible.
3+ of any polygons with 7+ sides is impossible.