Dimension of $\Lambda ^k (V)$

Question

In Calculus on Manifolds by Spivak theorem 4-5 is as follows:

The set of all $$ \varphi _{i_1} \wedge \dots \wedge \varphi _{i_k} \quad 1\leq i_1<i_2<\dots<i_k\leq n$$ is the basis for $\Lambda ^k (V)$, which therefore has dimension $\binom{n}{k}$.

Why do we have the condition $ 1\leq i_1<i_2<\dots<i_k\leq n$ (strict order) ?

Answer 1

Because $\varphi_{i_1} \wedge \dots \wedge \varphi_{i_k}$ and $\varphi_{i_{j_1}} \wedge \dots \wedge \varphi_{i_{j_k}}$, where $(i_{j_1},\dots,i_{j_k})$ is a permutation of $(i_1,\dots,i_k)$, are linearly dependent. Further, since $\wedge$ is skew-symmetric, whenever two of the indices coincide your element is actually $0$.

Answer 2

Suppose that $n=2$. Then $\varphi_1\wedge\varphi_1=\varphi_2\wedge\varphi_2=0$. Furthermore, $\varphi_1\wedge\varphi_2=-\varphi_2\wedge\varphi_1$. Therefore, we only have to conseder $\varphi_1\wedge\varphi_2$ to get a basis.

The general case is similar.