Assume one of the points is the origin, the others are the vectors *a*_{1},...,*a*_{n+1}. If the pairwise distance is 1, then $a^2_i=1$ and $(a_i-a_j)^2=1$ (*i* < *j*)(scalar multiplication). This gives $(a_i,a_j)=\frac{1}{2}$. Now show that $a_1,\dots,a_{n+1}$ are linearly independent: if $\lambda_1,\dots,\lambda_{n+1}$ are scalars and $\sum \lambda_ia_i=0$, then scalar multiplication by $a_i$ gives $\lambda_i+\frac{1}{2}\sum_{j\neq i}\lambda_j=0$, then we have $\lambda_i=-\Lambda$ for each $i$ where $\Lambda=\lambda_1+\cdots+\lambda_{n+1}$, summing gives $\Lambda=0$ so each $\lambda_i=0$.