Prove that if vectors are independent then their cross product is not $0$ Let $V_1, V_2, ... V_{n-1} \in R^n$ be independent vectors. I want to prove that their cross product $V_1 \times V_2 \times ...\times V_{n-1} \ne 0$.
I know that the cross product is equal to the determinant of: $$
    \begin{pmatrix}
    e_1 & e_2 & \cdots & e_n \\
    V_{1,1} & V_{1,2} & \cdots & V_{1,n} \\
    \vdots & \vdots & \ddots \\
    V_{n-1,1} & V_{n-1,2} & \cdots & V_{n-1,n} \\
    \end{pmatrix}
$$  which is then equal to $$e_1\begin{vmatrix}
    V_{1,2} & V_{1,3} & \cdots & V_{1,n} \\
    \vdots & \vdots & \ddots \\
    V_{n-1,2} & V_{n-1,3} & \cdots & V_{n-1,n} \\
    \end{vmatrix} + \space... \space + e_n\begin{vmatrix}
    V_{1,1} & V_{1,2} & \cdots & V_{1,n-1} \\
    \vdots & \vdots & \ddots \\
    V_{n-1,1} & V_{n-1,2} & \cdots & V_{n-1,n-1} \\
    \end{vmatrix}$$ 
So it is enough to prove that one of those sub-determinants is not zero.
However when I to assume they all zero, so the rows are linearly dependent and to get to a contradiction I got stuck.
Any help here?
 A: If $V_1, V_2, \dots, V_{n-1}$ are independent in $\mathbb R^n$, there is an $n^{\text{th}}$ vector $V_n \in \mathbb R^n$ forming a basis with them.
Then we have
$$
   (V_1 \times V_2 \times \dots \times V_{n-1}) \mathbin{\boldsymbol{\cdot}} V_n = \det\begin{bmatrix}V_{n1} & V_{n2} & \cdots & V_{nn} \\ V_{11} & V_{12} & \cdots & V_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ V_{n-1,1} & V_{n-1,2} & \cdots & V_{n-1,n}\end{bmatrix}
$$
and this determinant is nonzero because $V_1, V_2, \dots, V_n$ are linearly independent. So $V_1 \times V_2 \times \cdots \times V_{n-1}$ can't be the zero vector, otherwise it could not have a nonzero dot product with $V_n$.
If you're not convinced that the dot product above is equal to the determinant, expand the cross product as you have done, then take the dot product with $V_n$. Since $$(a_1 e_1 + \dots + a_n e_n) \mathbin{\boldsymbol{\cdot}} b = (a_1 b_1 + \dots + a_n b_n),$$ you get the same expansion, but for the determinant with $V_n$ in it.
A: The other answer covers it, but here is an approach that uses the Riesz Representation theorem: define the linear functional 
$$\phi(x)=\det(V_1,\cdots, V_{n-1},x)$$ 
We apply Riesz to get a nonzero $z\in \mathbb R^n$ such that 
$$\phi(x)=\langle x,z\rangle$$ 
Then, 
$$\det(V_1,\cdots, V_{n-1},x)=\langle x,z\rangle$$ 
Now, 
$$\langle e_i,z\rangle=\det(V_1,\cdots, V_{n-1},e_i)$$
is the $i^{\text{th}}$ subdeterminant in your post. If each of these were zero then, 
$$\langle e_i,z\rangle =0;\ 1\le i\le n$$
and so $z=0$, a contradiction.
