Let $f: V \to V$ a linear map and $\dim V = n$. We have that $f^{\ast}: \Lambda^n(V) \to \Lambda^n(V)$ is necessarily a multiplication by some constant $c$. Prove that $c = \det f$.
My attempt. $\quad$ First, $\dim \Lambda^n(V) = 1$. Since $f^{\ast}$ is linear, then $f^{\ast}$ must be a multiplication by a constant.
If $\{e_1,...,e_n\}$ is a basis of $V$, then
$$f^{\ast}(\omega(v_1,...,v_n)) = \omega(f(v_1),...,f(v_n)).\tag{1}$$
Also,
$$\omega(v_1,...,v_n) = \det(a_{ij})\omega(e_1,...,e_n)\tag{2}$$
if $$v_i = \sum_j a_{ij}e_j.$$ So, $$f(v_i) = \sum_j a_{ij}f(e_j).$$ Thus, $$f^{\ast}\omega = \det(a_{ij})\omega(f(e_1),...,f(e_n)) = (\det f)\omega$$
Is this correct?