Justification for $\operatorname{dim} \wedge^k(V)$ is $n\choose k$ the alternating $k$-linear form. I was told that: $\operatorname{dim} \wedge^k(V)$ is $n\choose k.$ and I am trying to justify this by computing it for different dimensions of $V.$
If $\operatorname{dim}(V) = 1,$ then $v_1 \wedge v_1 = 0$. Therefore, $\operatorname{dim} \wedge^1(V)$ is $1\choose 1$
If $\operatorname{dim}(V) = 2,$ then we will have the following cases:
\begin{align*}
    v_1 \wedge v_1 &= 0\\
    v_1 \wedge v_2 &= -  v_2 \wedge v_1\\
     v_2 \wedge v_1 &= -  v_1 \wedge v_2\\
      v_2 \wedge v_2 &= 0
\end{align*}
But then I can not conclude what is my $k$ in this case and how many variables actually I am choosing as I have the second case is just the same as the third case. Could anyone help me in this please?
Also, I know that
\begin{align*}
    v_1 \wedge v_1 \wedge v_1 &= 0\\
    v_1 \wedge v_2 \wedge v_3 &= -  v_2 \wedge v_1 \wedge v_3\\
&= -  v_3 \wedge v_2 \wedge v_1\\
&= - v_1 \wedge v_3 \wedge v_2\\
&= v_2 \wedge v_3 \wedge v_1\\
&= v_3 \wedge v_1 \wedge v_2\\
     v_2 \wedge v_2 \wedge v_2  &= 0\\
      v_3 \wedge v_3  \wedge v_3 &= 0
\end{align*}
But I do not know how to translate this into $n \choose k.$ Could someone help me in this also?
Thanks in advance!
EDIT:
Note that I was told that my question maybe a duplicate of this If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$ but I do not agree with this as I need a small and concrete example to be calculated not an abstract procedure as in the previous link and the other one is asking about a specific case when $k<n$ in $k \choose n$ which is not my case.
 A: If $v_1, \dots, v_n$ is a basis for $V$ then for

*

*$k = 1$, we have $v_1, \dots, v_n$ as a basis for $\bigwedge^1V = V$ (definition)

*$k = 2$, we have $\{v_i \wedge v_j : i < j\}$ as a basis for $\bigwedge^2 V$

*$k = 3$, we have $\{v_i \wedge v_j \wedge v_l : i < j < l\}$ as a basis for $\bigwedge^3 V$.

and so forth. By definition, $k$ is the number of vectors you see in each expression. So $v_1$ is $k = 1$ since there is one vector in the expression; $v_1 \wedge v_2$ is $k = 2$ since there are two vectors, and so forth.
You said you understand that $v_i \wedge v_j = - v_j \wedge v_i$ and $v_i \wedge v_j \wedge v_l = v_j \wedge v_l \wedge v_i =$ etc. so I won't belabour that point.
Imagine $\dim V = 3$ and $k = 2$. Then a general vector in $V$ looks like $av_1 + bv_2 + cv_3$ and you can expand, for example
$$(av_1 + bv_2 + cv_3) \wedge v_2 = a (v_1 \wedge v_2) + b (v_2 \wedge v_2) + c(v_3 \wedge v_2) = a (v_1 \wedge v_2) - c (v_2 \wedge v_3)$$
Using the identities $v_i \wedge v_i = 0$ and $v_j \wedge v_i = - v_i \wedge v_j$.
If you expand $(av_1 + bv_2 + cv_3) \wedge (a'v_1 + b'v_2 + c'v_3)$ and collect terms, you will get some linear combination of $v_1 \wedge v_2, v_1 \wedge v_3, v_2 \wedge v_3$. (First expand one side, then the other.)
Next there is the combinatorial identity: $\#\{(i, j) : i < j\} = \#\{\{i, j\} : i \neq j \} = \binom{n}{2}$ because given a set of numbers like $\{5, 3\}$ there is exactly one way to create an increasing sequence $3 < 5$ out of it. Therefore increasing sequences of length $k$ and subsets of size $k$ are equinumerous.
A: How many totally antisymmetric tensors of rank $k$ exist in $\mathbb{R}^n$? We will take a combinatorial approach. Call the tensor $T_{i_1 i_2 ... i_k}$. If any of the $i_m$ are equal, then this component of $T$ is necessarily $0$, since interchanging any two indices flips the sign of their respective components. There are $k$ indices, each of which must take each of $1, ..., n$ distinct values (since the component of the tensor is determined as $0$ otherwise). So the amount of ways to choose a subset of size $k$ from $n$ is $^nC_k$. This means that the dimension of the $k$-th exterior power is $^nC_k$ (in $\mathbb{R}^n$). Hopefully this provides some intuition that you are looking for.
