Create an N-Dimensional Simplex (Tetrahedron)
Python 78 66 Bytes
lambda n:[i*[0]+[n]+(n+~i)*[0]for i in range(n)]+[n*[1+(n+1)**.5]]

Surely can be improved, especially at handling n=1```. (How is that even a simplex?) Just realized that's not necessary. Can probably be improved still ^^
Try it online!
[i*[0]+[1]+(n+~i)*[0]for i in range(n)] creates identity matrix. All points have distance sqrt(2) from each other. (thanks to Rod for improving)
Now we need a n+1-th point with the same distance to all other points. We have to choose (x, x, ... x).
Distance from (1, 0, ...  ) to (x, x, ... x) is sqrt((x-1)²+x²+...+x²). If we want an n dimensional simplex this turns out to be sqrt((x-1)²+(n-1)x²), as we have one 1 and n-1 0s in the first point. Simplify a bit: sqrt(x²-2x+1+(n-1)x²) = sqrt(nx²-2x+1)
We want this distance to be sqrt(2).
sqrt(2) = sqrt(nx²-2x+1)
2 = nx²-2x+1
0 = nx²-2x-1
0 = x²-2/n*x+1/n
Solving this quadratic equation (one solution, other one works fine, too):
x = 1/n+sqrt(1/n²+1/n) = 1/n+sqrt((n+1)/n²) = 1/n+sqrt(n+1)/n = (1+sqrt(n+1))/n
Put that in a list n times, put that list in a list and join with identity matrix.

-4 Bytes thanks to Alex Varga:
Multiply each vector by n. This changes the creation of the identity matrix to lambda n:[i*[0]+[n]+(n+~i)*[0] (same length) and gets rid of the division by n in the additional point, so it becomes n*[1+(n+1)**.5], saving two brackets and the /n.