Prove that $f^\ast\omega=\det f \cdot \omega$ Let $V$ be a vector space of dimension $n$ and  $f: V\to V$ a linear operator. I need to show that $f^\ast:\Lambda^n(V)\to \Lambda^n(V)$ is multiplication by $\det f$.
My try:
Since $\dim \Lambda^n(V)$ is $1-$dimensional and $f^\ast$ is linear, $f^\ast$ must be multiplication by a constant.
Let $\omega\in \Lambda^n(V)$ and let $A$ be the matrix of $f$ with respect to a basis $e_1,\dots,e_n$. Then $$f^\ast\omega(v_1,\dots,v_n)=\omega(f(v_1),\dots,f(v_n))=\det A \cdot \omega (e_1,\dots,e_n)$$
The first equality is definition and the second is 

4-6 $\ \ $ Theorem. $\ $ $\textit{Let }v_1,\ldots,v_n\textit{ be a basis for }V\textit{, and let }\omega\in\Lambda^n(V)\textit{. If }w_i=\sum\limits_{j=1}^n a_{ij}v_j\textit{ are }n\textit{ vectors in }V\textit{, then}$ $$\omega(w_1,\ldots,w_n)=\det(a_{ij})\cdot\omega(v_1,\ldots,v_n).$$

But I think this isn't what I what I need to obtain, namely $$f^\ast\omega(v_1,\dots,v_n)=\det A \cdot \omega (v_1,\dots,v_n)$$
Is there any mistake in my computations, or do I need some additional step?

The second equality should be
$$\omega(f(v_1),\dots,f(v_n))=\det A \cdot \omega (v_1,\dots,v_n).$$
Let $w_i = f(v_i)$, we have
$$
w_i  = f(v_i) = f(\sum_jv_{ji} e_j) = \sum_jv_{ji}f(e_j) = \sum_k\big(\sum_j A_{kj} v_{ji}\big)e_k = \sum_kw_{ki}e_k,
$$
Therefore by applying the theorem twice,
\begin{align}
\omega(w_1,\dots,w_n) &= \det [w_{ki}] \cdot \omega(e_1,\dots,e_n) \\ &=\det ([A_{kj}v_{ji}])  \cdot \omega(e_1,\dots,e_n)\\ &= \det A \cdot  \det [v_{ji}] \cdot \omega(e_1,\dots,e_n) \\ &= \det A  \cdot \omega(v_1,\dots,v_n)
\end{align}
