# Prove that determinant of transpose is the same as determinant using wedge product

Let V a vector space of finite dimension. Let dim $V$ = n.

I know that for an operator $f:V\to V$ I have a map $\bigwedge^nf:\bigwedge^nV\to\bigwedge^nV$ such that: $$\bigwedge^nf(v_1\wedge\dots\wedge v_n) = (fv_1\wedge\dots\wedge fv_n) =det f(v_1\wedge\dots\wedge v_n)$$ I want to prove that $detf = detf^*$ (where $f^*:V^*\to V^*$ defined by $f^*(\phi) = \phi \circ f$). I tried to think of $\bigwedge^nf$ being an isomorphism (since is multiplication by a number, determinant of f) and somehow connecting it to the transpose, but I don't know how.