How can one compute directly the dimension of $\Lambda^k (V)$ from this definition? Let $V$ be a vector space, and
$$\Lambda (V):=(T V)/I$$
be one of the usual (well-known) definitions of the exterior algebra. I am struggling on proving directly that $\{e_I\}$, where $I$ is any multi-index of increasing $k$ indices is a basis of $\Lambda ^k V$ as a vector space, where $\Lambda ^k V$ is by definition the vector space spanned by $\{e_J\}$, $J$ being any multi-index of $k$ indices. Every proof of this fact that I've seen is by the definition as multilinear alternating maps, and relies on taking dual bases etc, and I can't adapt it to this case.

Let $e_1,\ldots,e_n$ be a basis for $V$. I claim that $e_{i_1}\cdots e_{i_k}$ with $i_1<\cdots < i_k$ is a basis for $\wedge^k (V)$. Indeed, suppose that a sum $$\sum \lambda_I e_I=0$$
let $J$ be the unique subset of $[n]$ such that $I\cap J=\varnothing$ and $I\cup J=[n]$. Then taking the product with $e_J$ gives $$\lambda_I e_I\wedge e_J=0$$
so $\lambda_I=0$. Indeed, since $I$ is different from all the other $I'$, then $J$ must intersect $I'$ and $e_I'e_J=0$. As an example, consider $\lambda x y+\mu y z=0$, in $\wedge(x,y,z)$. Then taking the product with $x$ gives $\mu xyz=0$. 
