Definition: Let $\mathbb R^n$ be the $n$-dimensional real vector space. An exterior $k$-form call any skew-symmetric tensor on $\mathbb R^n$ of rank $k$. Denote the set of exterior $k$-forms by $E^k$.
Now $w\in E^k$ if and only if $w$ is poly-linear in all its arguments and $$\forall i\in\overline {1,n}\qquad w(c_1,\dots c_i,c_{i+1},\dots, c_k)=-w(c_1,\dots c_{i+1},c_i,\dots, c_k)$$
Definition: Define the exterior product "$\land$" on $1$-forms by $$w_1\land\dots\land w_k(c_1,\dots, c_k)=\begin{vmatrix} w_1(c_1) & w_2(c_1) & \dots & w_k(c_1)\\ w_1(c_2) & w_2(c_2) & \dots & w_k(c_2)\\ \vdots & \vdots & \ddots &\\ w_1(c_n) & w_2(c_n) & \dots & w_k(c_n)\\ \end{vmatrix}$$
Proposition: Assume $x_1,\dots, x_n$ is a basis of $(\mathbb R^n)^*$, the dual of $\mathbb R^n$. Then the set of basic $k$-forms $B=\{x_{i_1}\land\dots\land x_{i_k}\quad :\quad 1\leq i_1<i_2<\dots i_k\leq n \}$ is a basis of $E^k$ (considered as a real vector space with pointwise addition).
Attempt: Assume $e_1\dots e_n$ is the dual basis to $x_1,\dots, x_n$. To see that $B$ spans $E^k$ it suffices to see that any form $w$ is a linear combination $w=\sum_I a_Ix_I$. Here $I$ spans the increasing multi-indexes, and if $I=i_1,\dots ,i_k$ then $x_I:=x_{i_1}\land\dots\land x_{i_k}$, and $a_I:=w(x_{i_1},\dots,x_{i_k})$. The above is apparent from $i_l\not=i_m\Rightarrow x_i(e_j)=0$ and from the definition of the exterior product.
To establish linear independence I tried assuming to the contrary. Then I expressed an arbitrary element of the supposed basis as a linear combination of the others. I imagine evaluating both sides at some carefully chosen vectors should yield a contradiction. May I get help with the details?
N.B. I want to avoid looking at the exterior product as an alternation of the tensor product. Also, I read Rudins proof, but I cannot interpret it in terms of my definitions.
To establish linear independence I had expressed one of the basic form in terms of the others. That is, if $J$ spans the increasing indexes, then
$$x_I=\sum_Ja_Jx_J$$
Evaluating $x_I(e_I)$ yealds $1$ but the right hand side gives $0$, a contradictoin. Therefore, the basic forms are linearly independent.