Proof of the Laplace formula using alternating tensors. I am reading "Differential Forms with applications to the physical sciences" of Harley Flanders. I came across the proof of the Laplace formula, there were some details left to the reader, I would like to know if my proof is correct.
Let $\{a_{i,\,j}\}_{i=1,\dots,n,\,j=1\dots n}$  be an $n\times n$ matrix. For an order $p-$tuple $H=\{h_{1},\dots,\,h_{p}\}$ , set
$$b_{H}=\begin{vmatrix}
a_{1,\,h_{1}}&& \dots && a_{1,\,h_{p}}\\
\vdots && && \vdots \\
a_{p,h_{1}} && \dots && a_{p,h_{p}}
\end{vmatrix}.$$
Set $p+q=n.$ For $K=\{k_{1},\,\dots\, k_{q}\},$ set
$$c_{K}=\begin{vmatrix}
a_{p+1,\,k_{1}}&& \dots && a_{1,\,k_{q}}\\
\vdots && && \vdots \\
a_{n,k_{1}} && \dots && a_{n,k_{q}}
\end{vmatrix}.$$
I will indicate with $H^{c}$ the ordered $q-$tuple complementar of $H$. Set $\alpha_{i}=\sum_{j=1}^{n}a_{i,\,j}\,\sigma^{j},$ where $(\sigma^{j})$  is a base of $\mathbb{R}^{n}.$
First we prove by induction that
$$ \alpha_{1}\wedge\,\dots\,\wedge \alpha_{p}=\sum_{H} b_{H}\,\sigma^{H}$$
and
$$\alpha_{p+1}\wedge\,\dots\, \wedge \alpha_{n}=\sum_{K}c_{K}\,\sigma^{K}.$$
The case $p=1$ is trivial. If $p>1$ we have (I am not really sure about the third equality)
$$ \begin{align*}
\alpha_{1}\wedge\,\dots\,\wedge \alpha_{p} =\left(\sum_{j=1}^{n} a_{j,1}\, \sigma^{j}\right)\wedge \, \left( \alpha_{2}\,\wedge \dots \, \wedge \alpha_{p} \right)\\
=\sum_{j=1}^{n}\,a_{j,1}\,\sigma^{j} \wedge \sum_{\bar{H}}b_{\bar{H}}\,\sigma^{\bar{H}}=\sum_{i=1}^{n}a_{h_{i},1}\,\sigma^{h_{i}}\wedge b_{\hat{H}_{i}}\, \sigma^{\hat{H}_{i}}\\
=\sum_{i=1}^{n} a_{1,h_{i}}\,b_{\hat{H}_{i}}\,(-1)^{h_{i}-1}\,\sigma^{H}=\sum_{H}\, b_{H}\,\sigma^{H}.
\end{align*}$$
Where the last equality is by the definition of the determinant.
Remark: the notation $\hat{H}_{i}$ means the ordered $p-1-$tuple obtained by $H$ subtracting $h_{i}.$
The other formula
$$\alpha_{p+1}\wedge\,\dots\, \wedge \alpha_{n}=\sum_{K}c_{K}\,\sigma^{K}$$
is proven similarly.
So the Laplace formula is readily proven :
$$ 
\begin{align*}
\det(A)\,\sigma^{1}\wedge \dots \wedge \sigma^{n}=\left(\alpha_{1}\wedge \dots \wedge \alpha_{p}\right)\wedge \left( \alpha_{p+1}\wedge \dots \alpha_{n} \right)=\sum_{H,K} b_{H}\,c_{K}\,\sigma^{H}\wedge \sigma^{K}
\end{align*}
$$
But
$$
\sigma^{H}\wedge \sigma^{K}=\begin{cases}
0 \quad K\neq H^{c}\\
sgn(H,H^{c}) \quad K=H^{c}
\end{cases}
$$
In conclusion
$$ 
\begin{align*}
\det(A)\,\sigma^{1}\wedge \dots \wedge \sigma^{n}=\sum_{H,K} b_{H}\,c_{K}\,\sigma^{H}\wedge \sigma^{K}=\left( \sum_{H} sgn(H,H') a_{H}\,b_{H'} \right) \cdot \sigma_{1}\wedge \dots \sigma_{n}
\end{align*}
$$
Remark: obviously the notation $sgn(H,H')$ stands for the sign of the permutation
$$ \begin{pmatrix}
1 &2 & \cdots &p& p+1 & \dots & n\\
h_{1} & h_{2} & \cdots & h_{p}& k_{1} & \cdots &k_{q}.
\end{pmatrix}
$$
 A: First, it appears you are trying to prove the classical Laplace expansion formula for an $n\times n$ real matrix $A=(a_{ij})$ along the first $p$ rows, which can be written more explicitly as
$$\det A=\sum_{h_1<\cdots<h_p}\mathrm{sgn}(h_1,\ldots,h_p)A_{1\cdots p}^{h_1\cdots h_p}A_{p+1\cdots n}^{h_{p+1}\cdots h_n}\tag{1}$$
where $h_{p+1}<\cdots<h_n$ is the complementary ordered $(n-p)$-tuple of $h_1<\cdots<h_p$,
$$\mathrm{sgn}(h_1,\ldots,h_p)=(-1)^{\sum_{i=1}^p(h_i-i)}\tag{2}$$
and $A_{i_1\cdots i_k}^{j_1\cdots j_k}$ denotes the $k\times k$ minor determinant of $A$ from rows $i_1<\cdots<i_k$ and columns $j_1<\cdots<j_k$. (Note (1) is equivalent to your final outer equation, just using slightly more explicit notation; in particular, (2) is the sign of the permutation you describe.)
As in your attempted proof we can let $\sigma^1,\ldots,\sigma^n$ be a basis of $\mathbb{R}^n$ and $\alpha_i=\sum_{j=1}^n a_{ij}\sigma^j$.
Your attempted inductive proof of the expansion for $\alpha_1\wedge\cdots\wedge\alpha_p$ is problematic, even ignoring the typos. The right-hand side of your third equality doesn't make sense since $H$ (and hence $h_i$ and $\hat{H}_i$) is not defined there. However, it is true that
$$\alpha_1\wedge\cdots\wedge\alpha_p=\sum_{h_1<\cdots<h_p}A_{1\cdots p}^{h_1\cdots h_p}\sigma^{h_1\cdots h_p}\tag{3}$$
where $\sigma^{h_1\cdots h_p}=\sigma^{h_1}\wedge\cdots\wedge\sigma^{h_p}$, and this can be proved inductively using the Laplace cofactor expansion along a single row of a matrix (a simpler result which can be proved independently and which I assume you're familiar with).
As an illustrative example if $n=p=3$, then
$$\alpha_2\wedge\alpha_3=\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}\sigma^{12}+\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}\sigma^{13}+\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}\sigma^{23}$$
so it follows that
$$\begin{align*}
\alpha_1\wedge\alpha_2\wedge\alpha_3&=\Biggl(a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}\Biggr)\sigma^{123}\\
&=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}\sigma^{123}
\end{align*}$$
I trust you can generalize the reasoning here.
The rest of your proof looks right apart from typos.
