Exterior Power of a Vector Space Let $G$ be a finite group, $V$ an $n$-dimensional vector space over $\mathbb{C}$, and $\tau: G \rightarrow GL(V)$ a representation such that $\tau(g)$ has determinant 1 for all $g \in G$. Why is it that $\wedge^{n}(V) = \mathbb{C}$?
 A: An element $g\in G$ acts on $\Lambda^bV$ by multiplication by $\det\tau(g)$, which is $1$.
Indeed, let $B=\{v_1,\dots,v_n\}$ is a basis of $V$ and suppose $g\cdot v_i=\sum_{j=1}^ng_{i,j}v_j$ for each $i\in\{1,\dots,n\}$, so that the matrix $(g_{i,j})_{1\leq i,j\leq n}$ is the matrix of $\tau(g)$ with respect to the basis $B$. The element $\omega=v_1\wedge\cdots\wedge v_n\in\Lambda^nV$ is a non-zero element in that $1$-dimensional space, so that $\{\omega\}$ is a basis of $\Lambda^nV$, and 
\begin{align}
g\cdot\omega&=g\cdot v_1\wedge\cdots\wedge v_n\\
&=(g\cdot v_1)\wedge\cdots\wedge(g\cdot v_n)\\
&=\sum_{1\leq i_1,i_2,\dots,i_n\leq n}(g_{1,i_1}v_{i_1})\wedge\cdots\wedge (g_{n,i_n}v_{i_n})\\
&= \det(\tau(g)) \omega
\end{align}
Of course, one needs to justify the last equality. We start by rewriting the sum as $$\sum_{1\leq i_1,i_2,\dots,i_n\leq n}g_{1,i_1}\cdots g_{n,i_n}\;v_{i_1}\wedge\cdots\wedge v_{i_n}.$$ A term corresponding to a choice of $i_1$, $\dots$, $i_n$ in which we have repetitions is zero, so we can sum only over the choices of $i_1$, $\dots$, $i_n$ for which $(i_1,\dots,i_n)$ is a permutation of $(1,2,\dots,n)$, which we call $\pi$. If that is the case, then one should know that $$v_{i_1}\wedge\cdots\wedge v_{i_n}=\operatorname{sgn}(\pi)\;v_1\wedge\cdots\wedge v_n.$$ Doing this in all terms, we see that the sum is in fact equal to $\det(\tau(g))\;\omega$, in view of the definition of the determinant as a sum over permutations.
A: It is not strictly true that $\Lambda^n V=\mathbb C$, but only that $\Lambda^n V$ is a $1$-dimensional complex vector space.  
However the automorphism $\tau_g:V\to V$ has an $n$-th exterior power   which is an endomorphism
$\Lambda^n (\tau_g):\Lambda^n V\to \Lambda^n V$ of a  one-dimensional vector space and thus is multiplication by a  complex number . That number is $det (\tau_g)\in \mathbb C$.  
Note
At the level of elementary representation theory, it is permissible to be more down-to-earth and identify $V$ with $\mathbb C^n$, $GL(V)$ with $GL_n(\mathbb C)$, $\Lambda^n V$ with $\mathbb C$, etc.
However once you study more advanced material, that sort of confusion will lead to disaster: for example if $V$ is a rank-$n$ vector bundle, it is a terrible  mistake   to think that its determinant bundle is the trivial bundle with fibre $\mathbb C$.
So I encourage you to try and progressively learn the more canonical view as soon as possible.
