Prove vertices of a simplex are affinely independent

Given that the definition of a simplex $T$ is $x\in {\mathbb{R}}^{n}$ such that $x$ satisfies $n+1$ linear inequalities: $({u}_{k},x)<{c}_{k}$ for $k=1,\dots ,n+1$ (i.e. $T$ is the intersection of $n+1$ half spaces)

If we additionally impose the condition that $T$ must be bounded and nonempty, I was able to show that the ${k}^{\text{th}}$ vertex of $T$ can be defined ${v}_{k}$ as the unique solution to the system $({v}_{k},{u}_{j})={c}_{j}$ for all $j\ne k$The existence of such a solution comes from the fact that any size n subset of the ${u}_{k}$ are independent. Or else we could write ${u}_{k}$'s out as rows of a matrix and pick an arbitrarily large vector l in the nullspace. Adding that on to any existing $x\in T$ would show $T$ is unbounded since $x+l\in T$ since the inequalities $({u}_{k},x)<{c}_{k}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}({u}_{k},x+l)<{c}_{k}$.

It's intuitive to me that ${v}_{k}$'s must be affinely independent. It's also intuitive that the closure of $T$ should be the convex hull of all the ${v}_{k}$. In fact, I think either fact implies the other. But I'm unable to prove either of them. I was, however, able to prove that the convex hull of ${v}_{k}\subset T$.

Is there a simple proof that the vk are affinely independent?