I need your help to prove the following result:
Let $v_1,\ldots,v_{n+1}$ be vectors of $\mathbb{R}^n$ such that $\langle v_i,v_j \rangle < 0 \; \forall i \neq j$, where $\langle \cdot \, , \cdot \rangle$ is the standard dot product of $\mathbb{R}^n$. Prove that any set of $n$ of these vectors is a basis of $\mathbb{R}^n$.
Thanks in advance.
Remark: I changed the notation from $e_1,\ldots,e_{n+1}$ to $v_1,\ldots,v_{n+1}$ to avoid ambiguity with the canonical basis.
I think I found a demostration for the case $n=2$, which is at least a starting point.
Proof. Let $v_1,v_2,v_3$ be vectors of $\mathbb{R}^2$ such that
\begin{equation} \left.\begin{aligned} \langle v_1,v_2 \rangle &< 0 \\ \langle v_2,v_3 \rangle &< 0 \\ \langle v_1,v_3 \rangle &< 0 \end{aligned} \right\} \text{ $\Longrightarrow v_1 \neq v_2 \neq v_3 \neq \vec{0}.$ } \end{equation}
Let's suppose $v_1,v_2,v_3$ are pairwise linearly dependent, which means
\begin{aligned} v_2 &= \alpha v_1, \; \alpha \in \mathbb{R} \\ v_3 &= \beta v_2, \; \beta \in \mathbb{R} \\ v_3 &= \gamma v_1, \; \gamma \in \mathbb{R}. \end{aligned}
So we have
\begin{aligned} \langle v_1,v_2 \rangle &= \langle v_1,\alpha v_1 \rangle = \alpha \, \langle v_1,v_1 \rangle = \alpha \, \underbrace{ \|v_1\|^2 }_{> 0} < 0 \Longleftrightarrow \alpha < 0 \\ \langle v_2,v_3 \rangle &= \langle \alpha v_1,\gamma v_1 \rangle = \alpha \gamma \, \langle v_1,v_1 \rangle = \alpha \gamma \, \underbrace{ \|v_1\|^2 }_{> 0} < 0 \Longleftrightarrow \alpha \gamma < 0 \\ \langle v_1,v_3 \rangle &= \langle v_1,\gamma v_1 \rangle = \gamma \, \langle v_1,v_1 \rangle = \gamma \, \underbrace{ \|v_1\|^2 }_{> 0} < 0 \Longleftrightarrow \gamma < 0. \end{aligned}
But, if $\alpha < 0$ and $\gamma < 0$, it can't be $\alpha \gamma < 0$, so we have a contradiction, which means that $v_1,v_2,v_3$ are pairwise linearly independent. Finally, as $\mathrm{dim} \left( \mathbb{R}^2 \right) = 2$, any set of two vectors chosen among $v_1,v_2,v_3$ is a basis of $\mathbb{R}^2$.
$\blacksquare$
Ok, I got it. First of all, I want to thank Fimpellizieri, as I'm just actually going to write down more extensively what he/she suggested me in the comments.
Proof. Before starting let's notice that
$$\langle v_i,v_j \rangle < 0 \; \forall i \neq j \Longrightarrow v_k \neq \vec{0} \; \forall k \in \{1,\ldots,n+1\}.$$
Let's suppose that $n$ vectors among $v_1,\ldots,v_{n+1}$ are linearly dependent. For example, let's suppose, without loss of generality, that $v_1,\ldots,v_n$ are linearly dependent, i.e. $\exists \,(\lambda_1,\ldots,\lambda_n) \neq (0,\ldots,0)$ such as
$$\lambda_1 v_1 + \ldots + \lambda_n v_n = \vec{0}. \qquad \left(\small{1}\right)$$
Let's scalarly multiply by $v_{n+1}$ both members of $\left(\small{1}\right)$:
$$\langle \lambda_1 v_1 + \ldots + \lambda_n v_n, v_{n+1} \rangle = \langle \vec{0}, v_{n+1} \rangle,$$
i.e., for the linearity of the dot product,
$$\lambda_1 \, \underbrace{\langle v_1,v_{n+1} \rangle}_{< \, 0} + \ldots + \lambda_n \, \underbrace{\langle v_n,v_{n+1} \rangle}_{< \, 0} = 0.$$
This means that some coefficients $\lambda_i$ must be $>0$ and other ones $<0$. Let's suppose, without loss of generality, that
$$\lambda_1,\ldots,\lambda_k > 0, \quad \lambda_{k+1},\ldots,\lambda_n < 0, \qquad k< n.$$
Let's rewrite $\left(\small{1}\right)$ and separate the two categories of coefficients:
$$( \lambda_1 v_1 + \ldots + \lambda_k v_k ) + ( \lambda_{k+1} v_{k+1} + \ldots + \lambda_n v_n ) = \vec{0}$$
$\Longrightarrow ( \lambda_{k+1} v_{k+1} + \ldots + \lambda_n v_n ) = -( \lambda_1 v_1 + \ldots + \lambda_k v_k )$
$\Longrightarrow \langle (\lambda_1 v_1 + \ldots + \lambda_k v_k), (\lambda_{k+1} v_{k+1} + \ldots + \lambda_n v_n) \rangle = \\ \quad = \langle (\lambda_1 v_1 + \ldots + \lambda_k v_k), -(\lambda_1 v_1 + \ldots + \lambda_k v_k) \rangle = -\|\lambda_1 v_1 + \ldots + \lambda_k v_k\|^2. \qquad \left(\small{2}\right)$
On the other hand
$\langle (\lambda_1 v_1 + \ldots + \lambda_k v_k), (\lambda_{k+1} v_{k+1} + \ldots + \lambda_n v_n) \rangle = \\ = \lambda_1 \lambda_{k+1} \, \langle v_1,v_{k+1} \rangle + \ldots + \lambda_1 \lambda_n \, \langle v_1,v_n \rangle + \ldots + \lambda_k \lambda_{k+1} \, \langle v_k,v_{k+1} \rangle + \ldots + \lambda_k \lambda_n \, \langle v_k,v_n \rangle. \, \left(\small{3}\right)$
By putting together $\left(\small{2}\right)$ and $\left(\small{3}\right)$ we get
$\overbrace{\underbrace{\lambda_1 \lambda_{k+1}}_{<\,0} \, \underbrace{\langle v_1,v_{k+1} \rangle}_{<\,0}}^{>\,0} + \ldots + \overbrace{\underbrace{\lambda_1 \lambda_n}_{<\,0} \, \underbrace{\langle v_1,v_n \rangle}_{<\,0}}^{>\,0} + \ldots + \overbrace{\underbrace{\lambda_k \lambda_{k+1}}_{<\,0} \, \underbrace{\langle v_k,v_{k+1} \rangle}_{<\,0}}^{>\,0} + \ldots + \overbrace{\underbrace{\lambda_k \lambda_n}_{<\,0} \, \underbrace{\langle v_k,v_n \rangle}_{<\,0}}^{>\,0} \\ = \underbrace{-\|\lambda_1 v_1 + \ldots + \lambda_k v_k\|^2}_{<\,0},$
which is a contradiction. Consequently, $v_1,\ldots,v_n$ are linearly independent and, as $\mathrm{dim} \left( \mathbb{R}^n \right) = n$, they form a basis of $\mathbb{R}^n$. Finally, because of the arbitrariness of $v_1,\ldots,v_n$, any set of $n$ vectors chosen among $v_1,\ldots,v_{n+1}$ is a basis of $\mathbb{R}^n$.
$\blacksquare$
