Suppose $V$ is a finite-dimensional vector space over $\mathbb{C}$ and $T \in \mathcal{L}(V)$. Let $T^*$ be the operator on the dual vector space. Show that $\det(T) = \det(T^*)$.
My attempt :
We can choose a basis $e_1,e_2,\dots,e_n$ of $V$ such that $T$ is upper triangular. Then $\det(T)$ is the product of the diagonal entries of $T$. I am not sure how to proceed from here. I know that $T^*:V^* \rightarrow V^*$ is such that for $T^*(\phi) = \phi \circ T$ for all $\phi \in V^*$.
Any help would be tremendously appreciated.