Does there exist a higher-dimensional 5-sided "tetrahedron + 1"? The first shape is "0"-sided and is a point.  The next shape is a line segment and it's "1"-sided.  The next shape is a triangle and it's 3-sided.  The next shape is a tetrahedron and it's 4-sided.  Can we define some higher-dimensional 5-sided regular shape which is the next shape in this sequence?
You can reach the next shape by placing a point equidistant to all previous points in the next higher dimension.
 A: The simplex is the easiest polytope, existing within all dimensions. It is nothing but the pyramid on a simplicial base (of one dimension less). In fact, take any simplex, attach to every facet a copy of that simplex again, and fold those copies up into the next dimension, then those will all meet at a single point atop the central simplex, which then will become its base.
This very construction already shows by induction, that the D-dimensional simplex would have D+1 facets, which all are D-1 dimensional simplices. Thus I can be affirmative, the 4th dimensional simplex is a pentachoron, i.e. it is built from 5 cells, all being tetrahedra.
In fact, the element count of all subelements of any dimension would be given by the numbers of the Pascal triangle: eg. a triangle has 3 vertices and 3 sides; a tetrahedron has 4 vertices, 6 edges, and 4 faces; a pentachoron has 5 vertices, 10 edges, 10 triangles, and 5 tetrahedra; etc.
The constructive device given above surely requires that the circumradius of the lower dimensional simplex is less than one edge unit, in order to allow to build an unit-edged pyramid on top. Again by induction, using right that very construction of a D-dimensional simplex, you could derive that crucial circumradius formula being 
$$r=\sqrt{\frac{D}{2(D+1)}},$$
thus prooving its existance for every dimension.
--- rk
