Show that $\big\{A(v_{i_1}\otimes \cdots\otimes v_{i_k})\in V^{\wedge k}:i_1<\cdots<i_k \big\} $ is linearly independent

Since $V$ is finite-dimensional, we can view $V \approx (V^*)^*$. With this identification, view the $k$th tensor power of $V$ as the vector space of multilinear maps from $(V^*)^k$ to $\mathbb{R}$. Since $v_1, \dots, v_n$ is a basis of $V$, it is dual to some basis $\varepsilon_1, \dots, \varepsilon_n$ of $V^*$. Suppose that $\alpha = \sum_{I}a_IA(v_{i_1} \otimes \dots \otimes v_{i_k}) = 0$, where the sum is over all $I = (i_1, \dots, i_k)$ such that $i_1 < \dots < i_k$. We need to show that $a_I = 0$ for all $I$. We have $$\alpha(\varepsilon_{i_1}, \dots, \varepsilon_{i_k}) = \frac{1}{k!}a_{I}$$ since $A(v_{j_1} \otimes \dots \otimes v_{j_k})(\varepsilon_{i_1}, \dots, \varepsilon_{i_k}) = 0$ when $J = I$, and is $1$ when $J = I$. Hence $a_I = 0$ for all $I$.