If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$ 
If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$

I don't understand this result. If $k \lt n$ then how does the quantity $\binom {k} {n}$ make sense? Can anybody please help me in this regard?
Thanks for your time.
EDIT $:$ Let $V$ be a vector space. Then for any $n \in \Bbb N$ we can identify $V^{\times n}$ by $V^{\{1,2, \cdots, n \}}.$ Since $\{1,2, \cdots, n \}$ is an $S_n$-set it follows that $V^{\{1,2, \cdots, n \}}$ is an $S_n$-vector space. For any $\sigma \in S_n$ and any function $f : \{1,2, \cdots, n \} \longrightarrow V$ the action of $\sigma$ on $f$ is given by $(\sigma \cdot f) (i) = f (\sigma^{-1} (i)).$ This shows that $V^{\times n}$ is a $S_n$-vector space. Let $\mu$ be the action of $S_n$ on $V^{\times n}$ then for each $\sigma \in S_n$ the map $\otimes_V \circ \mu (\sigma, \cdot) : V^{\times n} \longrightarrow V^{\otimes n}$ will be a bilinear map. Then by universal property of tensor products there exists a unique bijective linear map $\rho (\sigma) : V^{\otimes n} \longrightarrow V^{\otimes n}.$ This gives a group homomorphism $\rho : S_n \longrightarrow GL(V^{\otimes n}).$ Hence $V^{\otimes n}$ is also a $S_n$-vector space where for any element $\sigma \in S_n$ and for any $v \in V^{\otimes n}$ the action of $\sigma$ on $v$ is given by $\sigma \cdot v : = \rho (\sigma) (v).$ If $\rho (\sigma) (v) = \left (\text {sign} (\sigma) \right ) v,$ for all $\sigma \in S_n$ and for some $v \in V^{\otimes n}$ then $v$ is said to be an alternating/anti-symmetric tensor. The collection of all alternating/anti-symmetric tensors forms a subspace of $V^{\otimes n}$ which is denoted by $\wedge^n V$ or $V^{\wedge n}.$
 A: We can define a projection from $V^{\otimes n}$ to $\wedge^n V$, the so called antisymmetrization map:
$$\pi:v\mapsto\frac{1}{n!}\sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\cdot\rho(\sigma)(v).$$
You can check for yourself that $\pi(v)$ is antisymmetric for all $v\in V^{\otimes n}$, and that $\pi(v)=v$ if $v$ is antisymmetric. So this is a surjective map $V^{\otimes n}\to\wedge^n V$.
Now consider the fact that surjective linear maps map generating sets to generating sets. In particular, they map bases to generating sets. Now let $B:=\{b_1,\dots,b_k\}$ be a basis of $V$. Then the set $\{c_1\otimes\dots\otimes c_n~\vert~c_i\in B\}$ is a basis of $V^{\otimes n}$. And thus $\{\pi(c_1\otimes\dots\otimes c_n)~\vert~c_i\in B\}$ is a generating set of $\wedge^n V$. Let's call $\pi(v_1\otimes\dots\otimes v_n)$ for $v_1,\dots,v_n\in V$ by another name, just for practicality:
$$v_1\wedge\dots\wedge v_n:=\pi(v_1\otimes\dots\otimes v_n).$$
This is the exterior product of $v_1,\dots,v_k$, and it has the nice property to be linear in each component and antisymmetric. All elements of $\wedge^n V$ can be written as linear combinations of such exterior products, and in particular, $\{c_1\wedge\dots\wedge c_n~\vert~c_i\in B\}$ is a generating set of $\wedge^n V$. It is not linearly independent, but you can choose a linearly independent subset: Throw away all exterior products containing basis vectors multiple times, since those are $0$. Also, among all exterior products containing the basis vectors $b_{i_1},\dots,b_{i_n}$, choose only one, since these are all multiples of each other due to antisymmetry. Once you did this, you have a linearly independent generating set (check this yourself). It consists of elements of the form $b_{i_1}\wedge\dots\wedge b_{i_n}$, where $b_{i_j}$ are all distinct and only one possible ordering is chosen. So among $k$ basis vectors, you choose $n$ distinct ones, without caring about the ordering. So you choose $n$ from $k$: $n$ choose $k$, or $\binom{k}{n}$ of them.
Also note, if $k<n$, there are no antisymmetric tensors except the $0$-tensor. This means the dimension is $0$ in that case. In accordance with the combinatorial definition of $\binom kn$, which makes it $0$ for $k<n$.
A: When $k<n$, the binomial coefficient ${{k}\choose{n}}$ is $0$. Also, when $k<n$ and $k = \dim V$, note that it is impossible to choose $n$ different basis vectors of $V$ to wedge together. As the space $\bigwedge^n V$ is spanned by all such exterior products with $n$ factors, and an exterior product is $0$ if the same basis vector appears in it twice, $\bigwedge^n V = \lbrace 0 \rbrace$, which has dimension $0 = {{k}\choose{n}}$.
A: You could try constructing a basis for $\bigwedge^n V$, that is proving the following result:
If $\{e_i\}_{i=1}^k$ is a basis for a vector space $V$ over $\mathbb{F}$, then
\begin{align*}
    \{e_{i_1} \wedge e_{i_2} \wedge \ldots \wedge e_{i_n} : 1 \leq i_1 < i_2 < \ldots < i_n \leq k\}
\end{align*}
is a basis for $\bigwedge^n V$ (where $\wedge$ for now is just a formal symbol).
From this it follows that $\dim \bigwedge^n V = \binom{k}{n}$. If $k<n$, by the definition of the binomial coefficient, the dimension of $\bigwedge^n V = 0$.
This makes perfect sense as the basis for $\bigwedge^nV$ are all of the possible combinations of $k$ basis vectors (in increasing subscript order). If we try to make a combination of $k>n$ basis vectors, there are zero such combinations in $\bigwedge^n V$, as all vectors in $\bigwedge^n V$ are alternating (i.e. do not have repeated elemetns, which can be shown by your definition of $\rho$).
