How generate n equidistant points in a n-1 dimensional space
I assume we are working in $\mathbb{R}^n$.

First of all, observe that one regular $n-$simplex effectively determines all the others. In fact, if $S_1, S_2$ are two sets of points in $\mathbb{R}^n$ that satisfy the regularity condition, then they can be obtained from each other by composing at most an isometry and an homothetic transformation of the affine space (the converse is also true).

Therefore, it is sufficient to construct an unitary $n-$simplex centered in the origin. We visualize each vertex of the simplex $v_1, v_2, \dots v_{n+1}$ as an element of the real $n$-dimensional vector space.

Consider two vertices $v_1, v_2$ of the simplex, let $O$ be the origin and $\pi$ be the plane passing through $v_1, v_2, O$. The angle $\vartheta =\angle v_1 O v_2$ is exactly $\arccos (-1/n)$. To prove that, we observe that:

$$
0 = \left \| \sum_{i} v_i \right \|^2 = (n+1) + 2 \cos\vartheta \binom{n+1}{2} 
$$

We deduce that the following hold:


$\forall i \quad \| v_i \| = 1$
$\forall i \neq j \quad \left \langle v_i, v_j\right \rangle = -1/n$


We may assume without loss of generality that $v_1, v_2$ lie on the same line, that $v_3$ lies on the same plane as the others and so on (in general, that for each $k$, $v_1, v_2, \dots v_k$ lie on the same subspace of $\mathbb{R}^n$ with dimension $k$). Therefore, we may write the vectors by coordinates as follows:

$$
\begin{array}{r c l}
\begin{array}{r}
v_1 \\ v_2 \\ \cdots \\ v_{n+1}
\end{array} &amp;
\begin{array}{r}
= \\ = \\ \\ =
\end{array} &amp;
\begin{array}{c c c c c c}
(x_{1,1} &amp; 0 &amp; 0 &amp; 0 &amp; \dots &amp; 0)^T \\ 
(x_{2,1} &amp; x_{2,2} &amp; 0 &amp; 0 &amp; \dots &amp; 0)^T \\
\\
(x_{n+1, 1} &amp; x_{n+1, 2} &amp; x_{n+1, 3} &amp; x_{n+1, 4} &amp; \dots &amp; x_{n+1, n+1})^T \\ 
\end{array}
\end{array}
$$

The first equation uniquely determines $x_{1,1}$ and the second all the $x_{m,1}$. Now we use again the first to compute $x_{2,2}$ and with the second we determine all the remaining $x_{2, m}$.

Continuing the procedure in a similar fashion computes the coordinate values of all the vertices.