All the vector spaces I will consider in the following could be thought to have a base field of characteristic $ 0 $ (just to be safe).
I know that if $ A\colon V\to W $ is a linear mapping between to vector spaces $ V $ and $ W $ and $ k\geqq 0 $ is an integer then $$ \Bigl(\bigwedge\nolimits^k A\Bigr)(v_{j_1}\wedge\dots \wedge v_{j_k}) = \sum_{i_1 < \dots < i_k}\det(A_{IJ}) w_{i_1}\wedge\dots \wedge w_{i_k} $$ where $ I = (i_1,\dots,i_k) $ and $ J = (j_1,\dots,j_k) $ are ordered multi-indices and $ A_{IJ} $ is the submatrix of the matrix $ A $ of the linear mapping $ A $ with respect to the bases $ \{v_1,\dots,v_n\} $ and $ \{w_1,\dots,w_m\} $ respectively of $ V $ and $ W $.
Now, take a vector space $ V $ and two bases $ \{v_1,\dots,v_n\} $ and $ \{w_1,\dots,w_n\} $ of $ V $. Define an automorphism $ A\colon V\to V $ by $ A(v_j) = w_j $. Then $$ w_{i_1}\wedge\dots \wedge w_{i_k} = A(v_{i_1})\wedge\dots \wedge A(v_{i_k}) = \Bigl(\bigwedge\nolimits^k A\Bigr)(v_{i_1}\wedge\dots \wedge v_{i_k}) $$ so by the previous result $$ w_{i_1}\wedge\dots \wedge w_{i_k} = \sum_{j_1 < \dots < j_k}\det(A_{JI})w_{j_1}\wedge\dots \wedge w_{j_k} $$ but then $$ \det(A_{JI}) = \delta_{JI} $$ where $$ \delta_{JI} = \begin{cases} 1 & \text{if $ J = I $}\\ 0 & \text{otherwise} \end{cases} $$ because the $ w_{j_1}\wedge\dots \wedge w_{j_k} $ are linearly independent.
This result feels suspicious. Have I done something illegal?
Everything is fine. With respect to the bases $\{v_1, \dots, v_n\}$ and $\{w_1, \dots, w_n\}$ the matrix of $A$ is the identity matrix, and the minors are indeed $\det A_{JI} = \begin{cases}1, & I = J \\ 0, & \text{otherwise.}\end{cases}$