Let $v_1, \ldots , v_{s+1}$ be vectors in $\mathbb R^n$. Suppose $v_i^T \cdot v_j < 0$ for $i \neq j$. Prove $v_i, \ldots, v_s$ are independent. 
Let $v_1, \ldots , v_{s+1}$ be vectors in $\mathbb R^n$. Suppose $v_i^T \cdot v_j < 0$ for $i  \neq j$. Prove $v_i, \ldots, v_s$ are independent.

Suppose $c_1 \cdot v_1 + c_2 \cdot v_2 + \ldots + c_s \cdot v_s = 0$. 
$v_{s+1}^T \cdot (c_1 \cdot v_1 + c_2 \cdot v_2 + \ldots + c_s \cdot v_s) = 0 = c_1 \cdot (v_{s+1}^T \cdot v_1) + c_2 \cdot (v_{s+1}^T \cdot v_2) + \ldots + c_s \cdot (v_{s+1}^T \cdot v_s) = 0$. 
I don't know how to go on from here. Can someone provide some hints?
Picture of the exercise:

 A: Suppose $c_1\cdot v_1 + ... +c_s\cdot v_s =0$.
We can separate the $c_i$ into non-negative and negative and rewrite as:
$\sum _{i} s_{i} v_i=\sum_{j}t_j v_j$ where $s_i, t_j \geq 0$.
Let $u=\sum_{i}s_i v_i$. Then $(u,u)=\sum_{i j} s_i t_j (v_i,v_j)\leq 0$.
This means that $u=0$.
Now we have that $0=(u,v_{s+1})=\sum_{i}s_i(v_i,v_{s+1})$.
Since $(v_i,v_{s+1})<0$, this means that $s_i=0$ for all $i$.
Similarly, one can show that $t_j=0$ for all $j$.
This means that $c_k=0$ for all $1\leq k\leq s$ which proves that $v_1,...,v_s$ are independent.
A: 
If you have $s$ linearly dependent vectors $v_i, 1 \leq i \leq s$ satisfying the scalar product hypothesis, then for any vector $v_{s+1}$ there must be a $j \leq s$ so that $v_j\cdot v_{s+1} > 0$.

This is another formulation of the same problem, and it is the one I intend to prove by induction. The induction step turned out to be a small wall of text, so here is a small summary:

First I rotate the coordinate axes in a way so that determining the sign of $v_i\cdot v_{s + 1}$ from coordinate inspection will be easy. Then I point out that there must be a vector $v_j$ so that $v_j \cdot v_{s+1} > 0$.

Now for the real proof.
The claim is obviously true for $s = 1$. (Remember that the only linearly dependant set of one vector is the set containing the zero vector.)
Assume the claim is true for $s-1$, and let $v_1, \ldots, v_s$ be vectors satisfying the negative salar product hypothesis. For convenience purposes, assume that $\operatorname{span}(v_1, v_2, \ldots, v_s) \subseteq \Bbb R^{s-1}$. We then have that $\operatorname{span}(v_1, v_2, \ldots, v_{s-1}) = \Bbb R^{s-1}$ by the induction hypothesis (take away any one of those vectors, and the rest are linearly independent, so they must span an $s-1$-dimensional space).
We can rotate $\Bbb R^{s-1}$ so that the coordinates of $v_i, i < s$ are nonzero only for the first $i$ coordinates, and that the $i$th coordinate is positive.  (We're rotating the coordinate axes so that the first axis align with $v_1$, the first plane contains $v_2$ in its second quadrant, and so on. This changes no scalar products).
Note that it might be impossible, because of orientation, to get the last coordinate of $v_{s-1}$ positive by a rotation that keep the other vectors. But then either swap the indices of $v_s$ and $v_{s-1}$, or allow the rotation to momentarily take us out of $\Bbb R^{s-1}$, and it will work out. Also worth noting is that this alignment of the coordinate system is the only place I use the induction hypothesis; I need the first $s-1$ vectors to be linearly independent in order to define this rotation nicely. It is, however, not strictly necessary.
The scalar product hypothesis then dictates that all other coordinates of $v_i$ have to be negative: Since $v_1\cdot v_i < 0$, $v_i$ must have first coordinate negative. Since this also goes for $v_2$, that means that for $v_2 \cdot v_i$ to be negative, $v_i$'s second coordinate must also be negative, and so on. This means that the coordinates of $v_s$ must all be negative.
Now, let's finally take a look at the vector $v_{s+1} \in \Bbb R^n$. It must have some coordinate representation in our newly rotated space. If the coordinates with index $\leq s-1$ are all negative, we have $v_s \cdot v_{s+1} > 0$. Otherwise, let $j$ be the index of the first positive coordinate. Then $v_j \cdot v_{s+1} > 0$. QED.
A: The statement is false. Pick $v_1 = [1, 0]$ and $v_2 = [-1, 0]$.  Perhaps the statement was supposed to say that $v_i^T \cdot v_j = 0$ for $i \ne j$? 
As noted in the comments, I had misread the statement, thinking it was claiming that all vectors were independent, rather than "all but the last one." 
