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?