INCOMPLETE: Signs are incorrect. Feel free to edit!

Here is a proof of $\left( \ast\ast\right) $. As you correctly admitted, part of the problem is to define the signs that enter into $\operatorname*{Cof}A$. Let me actually start from scratch and introduce all notations that will be needed:

If $U\subseteq\left[ n\right] $ and $V\subseteq\left[ d\right] $, then we let $A_{U,V}$ denote the submatrix of $A$ obtained by removing all rows except for those in $U$ and all columns except for those in $V$. (In more rigorous terms: We let $A_{U,V}$ be the matrix $\left( a_{u_{i},v_{j}}\right) _{1\leq i\leq\left\vert U\right\vert ,\ 1\leq j\leq\left\vert V\right\vert }$, where $A$ is written in the form $\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq d}$, and where the sets $U$ and $V$ are written in the forms $U=\left\{ u_{1}<u_{2}<\cdots<u_{\left\vert U\right\vert }\right\} $ and $V=\left\{ v_{1}<v_{2}<\cdots<v_{\left\vert V\right\vert }\right\} $.)

Define an element $\operatorname*{Det}A\in\wedge^{n-d}\left( \mathbf{k} ^{n}\right) $ by $\operatorname*{Det}A=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}$.

$\left( \operatorname*{Cof}A\right) \left( f_{i}\right) =\sum _{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}$.

$=\sum_{i\in\left[ d\right] }a_{k,i}\underbrace{\left( \operatorname*{Cof} A\right) \left( f_{i}\right) }_{\substack{=\sum_{\substack{J\subseteq \left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}\\\text{(by the definition of }\operatorname*{Cof}A\text{)}}}$

$=\sum_{i\in\left[ d\right] }a_{k,i}\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i} \det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus \left\{ i\right\} }\right) e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i} a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

(1) $=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

$+\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\notin J}}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$.

If $J$ is a finite set of integers, and if $k$ is an integer, then $\#\left( J<k\right) $ shall denote the number of all elements of $J$ that are smaller than $k$.

(2) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i} \det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus \left\{ i\right\} }\right) =\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) $

(by Laplace expansion, applied to the matrix $A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }$, with respect to the $\#\left( \left[ n\right] \setminus J<k\right) +1$-th row of this matrix (which used to be the $k$-th row of $A$)).

(3) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i} \det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus \left\{ i\right\} }\right) =0$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\underbrace{\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) } _{\substack{=\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) \\\text{(by (2))}}}e_{J}$

$+\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\notin J}}\underbrace{\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) } _{\substack{=0\\\text{(by (3))}}}e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) e_{J}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1} \underbrace{\det\left( A_{\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) \right) \cup\left\{ k\right\} ,\left[ d\right] }\right) }_{\substack{=\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) \\\text{(since }\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) \right) \cup\left\{ k\right\} =\left[ n\right] \setminus I\text{)}}}\underbrace{e_{I\cup \left\{ k\right\} }}_{\substack{=\left( -1\right) ^{\#\left( I>k\right) }e_{I}\wedge e_{k}\\\text{(because the sign of the permutation}\\\text{that takes }e_{I\cup\left\{ k\right\} }\text{ to }e_{I}\wedge e_{k}\text{ is }\left( -1\right) ^{\#\left( I>k\right) }\text{)}}}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) \left( -1\right) ^{\#\left( I>k\right) }e_{I}\wedge e_{k}$

(here, we have substituted $I\cup\left\{ k\right\} $ for $J$ in the sum, because every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying $k\in J$ must have the form $I\cup\left\{ k\right\} $ for some $k$-element subset $I$ of $\left[ n\right] $ satisfying $k\notin I$)

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\underbrace{\left( -1\right) ^{\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1+\#\left( I>k\right) }}_{\substack{=\left( -1\right) ^{n-d+k}\\\text{(since }\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1+\#\left( I>k\right) \\\equiv\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1-\#\left( I>k\right) \\=\left\vert I\right\vert +k=n-d+k\operatorname{mod} 2\text{)}}}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{n-d+k}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

$=\left( -1\right) ^{n-d+k}\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$.

$\underbrace{\operatorname*{Det}A}_{=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}}\wedge\underbrace{w}_{=e_{k}}$

$=\left( \sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\right) \wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

(here, we have removed all addends which satisfy $k\in I$, because all such addends contain the vanishing factor $e_{I}\wedge e_{k}=0$), we should obtain $\left( \left( \operatorname*{Cof}A\right) \circ A^{T}\right) \left( w\right) =\left( \operatorname*{Det}A\right) \wedge w$, and thus Theorem 1 should follow. But the signs don't match, and I am out of time. Sorry :(

Here is a proof of $\left( \ast\ast\right) $. As you correctly admitted, part of the problem is to define the signs that enter into $\operatorname*{Cof}A$. Let me actually start from scratch and introduce all notations that will be needed:

If $U\subseteq\left[ n\right] $ and $V\subseteq\left[ d\right] $, then we let $A_{U,V}$ denote the submatrix of $A$ obtained by removing all rows except for those in $U$ and all columns except for those in $V$. (In more rigorous terms: We let $A_{U,V}$ be the matrix $\left( a_{u_{i},v_{j}}\right) _{1\leq i\leq\left\vert U\right\vert ,\ 1\leq j\leq\left\vert V\right\vert }$, where $A$ is written in the form $\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq d}$, and where the sets $U$ and $V$ are written in the forms $U=\left\{ u_{1}<u_{2}<\cdots<u_{\left\vert U\right\vert }\right\} $ and $V=\left\{ v_{1}<v_{2}<\cdots<v_{\left\vert V\right\vert }\right\} $.)

If $S$ is a finite set of integers, then $\sum S$ shall mean the sum of the elements of $S$.

Define an element $\operatorname*{Det}A\in\wedge^{n-d}\left( \mathbf{k} ^{n}\right) $ by $\operatorname*{Det}A=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}$.

$\left( \operatorname*{Cof}A\right) \left( f_{i}\right) =\sum _{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i+\sum J}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}$.

$=\sum_{i\in\left[ d\right] }a_{k,i}\underbrace{\left( \operatorname*{Cof} A\right) \left( f_{i}\right) }_{\substack{=\sum_{\substack{J\subseteq \left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i+\sum J}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}\\\text{(by the definition of }\operatorname*{Cof}A\text{)}}}$

$=\sum_{i\in\left[ d\right] }a_{k,i}\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i+\sum J}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i+\sum J}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{\sum J}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

(1) $=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\sum J}\left( \sum _{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

$+\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\notin J}}\left( -1\right) ^{\sum J}\left( \sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e_{J}$

(here we have split the sum into two subsums, according to whether $k\in J$ or $k\notin J$).

If $J$ is a finite set of integers, and if $k$ is an integer, then $\#\left( J<k\right) $ shall denote the number of all elements of $J$ that are smaller than $k$. Similarly, $\#\left( J>k\right) $ shall denote the number of all elements of $J$ that are greater than $k$.

(2) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i+\#\left( \left[ n\right] \setminus J<k\right) +1}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) =\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) $

(by Laplace expansion, applied to the matrix $A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }$, with respect to the $\#\left( \left[ n\right] \setminus J<k\right) +1$-th row of this matrix (which used to be the $k$-th row of $A$)). Hence, every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying $k\in J$ satisfies

(3) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i} \det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus \left\{ i\right\} }\right) =\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) $

(this follows by multiplying both sides of (2) with $\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}$).

(4) $\sum_{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i} \det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus \left\{ i\right\} }\right) =0$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\sum J}\underbrace{\left( \sum _{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) }_{\substack{=\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) \\\text{(by (3))}}}e_{J}$

$+\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\notin J}}\left( -1\right) ^{\sum J}\underbrace{\left( \sum _{i\in\left[ d\right] }\left( -1\right) ^{i}a_{k,i}\det\left( A_{\left[ n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) }_{\substack{=0\\\text{(by (4))}}}e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\underbrace{\left( -1\right) ^{\sum J}\left( -1\right) ^{\#\left( \left[ n\right] \setminus J<k\right) +1}}_{=\left( -1\right) ^{\sum J+\#\left( \left[ n\right] \setminus J<k\right) +1}}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) e_{J}$

$=\sum_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\sum J+\#\left( \left[ n\right] \setminus J<k\right) +1}\det\left( A_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) e_{J}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1}\underbrace{\det\left( A_{\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) \right) \cup\left\{ k\right\} ,\left[ d\right] }\right) }_{\substack{=\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) \\\text{(since }\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) \right) \cup\left\{ k\right\} =\left[ n\right] \setminus I\text{)} }}\underbrace{e_{I\cup\left\{ k\right\} }}_{\substack{=\left( -1\right) ^{\#\left( I>k\right) }e_{I}\wedge e_{k}\\\text{(because the sign of the permutation}\\\text{that takes }e_{I\cup\left\{ k\right\} }\text{ to } e_{I}\wedge e_{k}\text{ is }\left( -1\right) ^{\#\left( I>k\right) }\text{)}}}$

(here, we have substituted $I\cup\left\{ k\right\} $ for $J$ in the sum, because every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying $k\in J$ must have the form $I\cup\left\{ k\right\} $ for some $\left( n-d\right) $-element subset $I$ of $\left[ n\right] $ satisfying $k\notin I$)

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) \left( -1\right) ^{\#\left( I>k\right) }e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\underbrace{\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1+\#\left( I>k\right) } }_{\substack{=\left( -1\right) ^{\sum I-\left\vert I\right\vert }\\\text{(since }\sum\left( I\cup\left\{ k\right\} \right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1+\#\left( I>k\right) \\\equiv\underbrace{\sum\left( I\cup\left\{ k\right\} \right) }_{=\sum I+k}+\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1-\underbrace{\#\left( I>k\right) }_{\substack{=\left\vert I\right\vert -\#\left( I<k\right) \\\text{(since }k\notin I\text{)}}}\\=\sum I+k+\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +1-\left\vert I\right\vert +\#\left( I<k\right) \\=\sum I+k+1-\left\vert I\right\vert +\underbrace{\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right) <k\right) +\#\left( I<k\right) }_{=k-1}\\=\sum I+k+1-\left\vert I\right\vert +k-1=\sum I+2k-\left\vert I\right\vert \\\equiv\sum I-\left\vert I\right\vert \operatorname{mod}2\text{)}}}\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum I-\left\vert I\right\vert } \det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$.

$\underbrace{\operatorname*{Det}A}_{=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}}\wedge\underbrace{w}_{=e_{k}}$

$=\left( \sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert } \det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\right) \wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }\det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

$=\sum_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert =n-d;\\k\notin I}}\left( -1\right) ^{\sum I-\left\vert I\right\vert } \det\left( A_{\left[ n\right] \setminus I,\left[ d\right] }\right) e_{I}\wedge e_{k}$

(here, we have removed all addends which satisfy $k\in I$, because all such addends contain the vanishing factor $e_{I}\wedge e_{k}=0$), we should obtain $\left( \left( \operatorname*{Cof}A\right) \circ A^{T}\right) \left( w\right) =\left( \operatorname*{Det}A\right) \wedge w$. This proves Theorem 1.