Determinant of the transpose via exterior products Let $V$ be a finite-dimensional vector space over $F$ and let $\tau:V \to V$ be a linear operator. Here's my definition of the determinant:

If $t:U \to U$ is a linear operator and $\dim(U)=n$ then $\det(t)$ is the unique number satisfying
  $$tu_1 \wedge \cdots \wedge tu_n = \det(t) u_1 \wedge \cdots \wedge u_n$$
  for all $u_1,\dots,u_n \in U$.

Define the transpose $\tau^T : V^* \to V^*$ of $\tau$ by $(\tau^Tf)(v)=f \tau v$. Is there a way of proving that $\det(\tau^T)=\det(\tau)$ without choosing a basis? It's clear that
$$\det(\tau) v_1 \wedge\cdots\wedge v_n = \tau v_1 \wedge\cdots\wedge \tau v_n$$
and
$$\det(\tau^T) f_1 \wedge\cdots\wedge f_n = \tau^T f_1 \wedge\cdots\wedge \tau^T f_n$$
for all $v_1,\dots,v_n \in V$ and $f_1,\dots,f_n \in V^*$, but how do I "join" them together?
 A: Define the "alternating" multilinear map
\begin{align}
(V^*)^n \times V^n &\to F \\
(f_1,\dots,f_n,v_1,\dots,v_n) &\mapsto \det(f_i(v_j))
\end{align}
where $\det(f_i(v_j))$ refers to the determinant of the matrix with its $(i,j)$ entry equal to $f_i(v_j)$. This gives us (details omitted) a nondegenerate pairing $\langle\cdot,\cdot\rangle:\bigwedge^n V^* \times \bigwedge^n V \to F$ with
$$(f_1 \wedge\cdots\wedge f_n,v_1 \wedge\cdots\wedge v_n) \mapsto \det(f_i(v_j)).$$
So for all $f_1,\dots,f_n \in V^*$ and $v_1,\dots,v_n \in V$,
\begin{align}
\det(\tau)\langle f_1 \wedge\cdots\wedge f_n, v_1 \wedge\cdots\wedge v_n \rangle &=
\langle f_1 \wedge\cdots\wedge f_n, \det(\tau)v_1 \wedge\cdots\wedge v_n \rangle \\
&= \langle f_1 \wedge\cdots\wedge f_n, \tau v_1 \wedge\cdots\wedge \tau v_n \rangle \\
&= \det(f_i\tau v_j) \\
&= \det((\tau^T f_i)v_j) \\
&= \langle \tau^T f_1 \wedge\cdots\wedge \tau^T f_n, v_1 \wedge\cdots\wedge v_n \rangle \\
&= \det(\tau^T)\langle f_1 \wedge\cdots\wedge f_n, v_1 \wedge\cdots\wedge v_n \rangle.
\end{align}
Therefore $\det(\tau)=\det(\tau^T)$.
