Highest number of equidistant points in $n$-dimensional space This seems a very basic problem so it's probably been tackled multiple times, but I am wondering how many equidistant* points can there be at an $n$-dimensional euclidean space (let's say $\mathbb{R}^n$ with the euclidean distance)
My conjecture is that, for $n$ dimensions, there can be exactly $n+1$ of those points. This is obvious for $n=1$. For $n=2$ we have the triangle and, for $n=3$, the tetrahedron. I cannot think of a way to arrange more points for $n \leq 3$ I think an extra dimension will always allow us to set a new point along that new axis, in the "center" of the previous figure (like going form a segment to a triangle and from a triangle to a tetrahedron)
So, am I right about my intuitions or do they break at some point (no pun intended)? Also, is it possible to have more than $n+1$ equidistant points?
So, am I right there?
*a set of points all separated the same distance from each other
 A: In fact those n+1 vertex polytopes of n-dimensional space with equal pairwise distances are called simplices (singular: simplex). This is, as those are the most simple polytopes within each dimension. For obvious reasons there will be no polytope within n-dimensional space with fewer vertices. Moreover those simplices all are regular polytopes.
The described construction by @AsuraPath is that of continued pyramidisation: take a vertex atop the simplex of one dimension less.
--- rk
A: For $n=2$, three equidistant points form an equilateral triangle (as you already know).
For $n=3$, let's see how you'd prove that four equidistant points form a tetrahedron:   Pick any three of the points.  These three points determine a plane, so now you've got three equidistant points in the plane, which we've already agreed must form an equilateral triangle.  Now draw a line through the center of that triangle and perpendicular to the plane in which the triangle lies.  It's quite obvious geometrically (and quite easy to prove if you just write down the algebra) that the fourth point lies on that line, and completes a tetrahedron.  
Now replace three and four with $n$ and $n+1$ and you have an inductive proof of your result.
