This is my proof without defining new notations.
Continuing from the induction hypothesis
$$\det{A}
=\sum_{j=1}^{n+1}(-1)^{1+j}[A]_{1,j}\det{A_{1,j}}
=\sum_{j=1}^{n+1}(-1)^{1+j}[A]_{1,j}\sum_{\sigma\in S_n}\text{sgn }\sigma\prod_{i=1}^{n}[A_{1,j}]_{i,\sigma(i)}$$
Denote $[n]=\{1, 2, ..., n\}$.
For any $\sigma\in S_n$,
since $\sigma$ is bijective,
let
$$i_1=\sigma^{-1}(1), i_2=\sigma^{-1}(2), ..., i_n=\sigma^{-1}(n).$$
Then $\{i_1, i_2, ..., i_n\}=[n]$
and $$\sigma(i_1)=1, \sigma(i_2)=2, ..., \sigma(i_n)=n.$$
As the following figure indicates.
$$\begin{matrix}
[n] & \sigma\in S_n & [n]\\
\hline
i_1 & \longrightarrow & 1 \\
i_2 & \longrightarrow & 2 \\
\vdots & \vdots & \vdots \\
i_n & \longrightarrow & n \\
\end{matrix}$$
Then
\begin{eqnarray*}
\det{A}
&=& \sum_{j=1}^{n+1}(-1)^{1+j}[A]_{1,j}\sum_{\sigma\in S_n}\text{sgn }\sigma\prod_{i=1}^{n}[A_{1,j}]_{i,\sigma(i)}\\
&=& \sum_{j=1}^{n+1}(-1)^{1+j}[A]_{1,j}\sum_{\sigma\in S_n}\text{sgn }\sigma\prod_{k=1}^{n}[A_{1,j}]_{i_k, \sigma(i_k)}\\
&=& \sum_{j=1}^{n+1}(-1)^{1+j}[A]_{1,j}\sum_{\sigma\in S_n}\text{sgn }\sigma\prod_{k=1}^{n}[A_{1,j}]_{i_k, k}\\
&=& \sum_{j=1}^{n+1}(-1)^{1+j}[A]_{1,j}\sum_{\sigma\in S_n}\text{sgn }\sigma\left(\prod_{k=1}^{j-1}[A_{1,j}]_{i_k, k}\prod_{k=j}^{n}[A_{1,j}]_{i_k, k}\right)\\
&=& \sum_{j=1}^{n+1}(-1)^{1+j}[A]_{1,j}\sum_{\sigma\in S_n}\text{sgn }\sigma\left(\prod_{k=1}^{j-1}[A]_{i_k+1, k}\prod_{k=j}^{n}[A]_{i_k+1, k+1}\right)\\
&=& \sum_{j=1}^{n+1}(-1)^{1+j}\sum_{\sigma\in S_n}\text{sgn }\sigma\left([A]_{1,j}\cdot \prod_{k=1}^{j-1}[A]_{i_k+1, k}\prod_{k=j}^{n}[A]_{i_k+1, k+1}\right)\\
&=& \sum_{j=1}^{n+1}(-1)^{1+j}\sum_{\sigma\in S_n}\text{sgn }\sigma \cdot [A]_{1,j}\cdot \underline{[A]_{i_1+1, 1}\cdot [A]_{i_2+1, 2}\cdots [A]_{i_{j-1}+1, j-1}}\cdot \\
&& \underline{[A]_{i_j+1, j+1}\cdot [A]_{i_{j+1}+1, j+2}\cdots [A]_{i_n+1, n+1}}\\
&=& \sum_{j=1}^{n+1}(-1)^{1+j}\sum_{\sigma\in S_n}\text{sgn }\sigma \cdot \underline{[A]_{i_1+1, 1}\cdot [A]_{i_2+1, 2}\cdots [A]_{i_{j-1}+1, j-1}}\cdot \\
&& [A]_{1,j}\cdot \underline{[A]_{i_j+1, j+1}\cdot [A]_{i_{j+1}+1, j+2}\cdots [A]_{i_n+1, n+1}}\\
\end{eqnarray*}
Consider a permutation $\tau_{\sigma}\in S_{n+1}$ as following
$$\begin{matrix}
[n+1] & \tau_{\sigma}\in S_{n+1} & [n+1]\\
\hline
i_1+1 & \longrightarrow & 1 \\
i_2+1 & \longrightarrow & 2 \\
\vdots & \vdots & \vdots \\
i_{j-1}+1 & \longrightarrow & j-1 \\
1 & \longrightarrow & j \\
i_j+1 & \longrightarrow & j+1 \\
\vdots & \vdots & \vdots \\
i_n+1 & \longrightarrow & n+1 \\
\end{matrix}$$
Then the equation above equals to
$$\det{A}=\sum_{j=1}^{n+1}(-1)^{1+j}\sum_{\sigma\in S_n}\text{sgn }\sigma\prod_{\ell=1}^{n+1}[A]_{\ell, \tau_{\sigma}(\ell)}.$$
Note that there is an one-to-one correspondence between $\sigma\in S_n$ and $\tau_{\sigma}\in S_{n+1}$ with $\tau_{\sigma}(1)=j$.
By the Lemma 2,
$\text{sgn }\tau_{\sigma}=(-1)^{1+j}\text{sgn }\sigma$.
Then
$$\det{A}=\sum_{j=1}^{n+1}\sum_{\substack{\tau\in S_{n+1}\\ \tau(1)=j}}\text{sgn }\tau\prod_{\ell=1}^{n+1}[A]_{\ell, \tau(\ell)}
=\sum_{\tau\in S_{n+1}}\text{sgn }\tau\prod_{\ell=1}^{n+1}[A]_{\ell, \tau(\ell)}.$$
Lemma 1.
If $\gamma\in S_{n+1}$ is
$$\begin{matrix}
[n+1] & \gamma\in S_{n+1} & [n+1]\\
\hline
1 & \longrightarrow & x_1 \\
2 & \longrightarrow & x_2 \\
\vdots & \vdots & \vdots \\
i & \longrightarrow & x_{i}\\
i+1 & \longrightarrow & x_{i+1}\\
\vdots & \vdots & \vdots \\
n+1 & \longrightarrow & x_{n+1}\\
\end{matrix}$$
Then $(x_i, x_{i+1})\gamma$ is
$$\begin{matrix}
[n+1] & \gamma\in S_{n+1} & [n+1]\\
\hline
1 & \longrightarrow & x_1 \\
2 & \longrightarrow & x_2 \\
\vdots & \vdots & \vdots \\
i & \longrightarrow & x_{i+1}\\
i+1 & \longrightarrow & x_{i}\\
\vdots & \vdots & \vdots \\
n & \longrightarrow & x_n \\
\end{matrix}$$
Lemma 2.
Back to our $\sigma$.
Consider $\sigma^{-1}\in S_n$.
We can define $\sigma^{-1}(n+1)=n+1$ to make it as an element in $S_{n+1}$.
That is,
$$\begin{matrix}
[n+1] & \sigma^{-1}\in S_{n+1} & [n+1]\\
\hline
1 & \longrightarrow & i_1 \\
2 & \longrightarrow & i_2 \\
\vdots & \vdots & \vdots \\
n & \longrightarrow & i_n \\
n+1 & \longrightarrow & n+1
\end{matrix}$$
By the Lemma 1, we can left-multiply a product of $m$ transpositions to make $i_1, i_2, ..., i_n, n+1$ in the right column in an increasing order.
In fact,
the product of these transpositions is $\sigma$.
Again, applying the Lemma 1 on $\tau_{\sigma}^{-1}\in S_{n+1}$ in the same way.
We can left-multiply $j-1$ transpositions to $\tau_{\sigma}^{-1}$ to move $1$ to the first element in the right column.
Then left-multiply $m$ transpositions to make $i_1+1, i_2+1, ..., i_n+1$ in the right column into an increasing order.
$$\begin{matrix}
[n+1] & \tau_{\sigma}^{-1}\in S_{n+1} & [n+1]\\
\hline
1 & \longrightarrow & i_1+1 \\
2 & \longrightarrow & i_2+1 \\
\vdots & \vdots & \vdots \\
j-1 & \longrightarrow & i_{j-1}+1 \\
j & \longrightarrow & 1 \\
j+1 & \longrightarrow & i_j+1 \\
\vdots & \vdots & \vdots \\
n+1 & \longrightarrow & i_n+1 \\
\end{matrix}$$
Suppose that
$s_m \cdots s_2 s_1 t_{j-1} \cdots t_2 t_1\tau_{\sigma}^{-1}=r_m\cdots r_2 r_1\sigma^{-1}=\varepsilon$,
where $s_m, ..., s_2, s_1, t_{j-1}, ..., t_2, t_1, r_m, ..., r_2, r_1$ all are transpositions
and $\varepsilon$ is the identity in $S_{n+1}$.
Therefore,
\begin{eqnarray*}
&&\text{sgn }(s_m \cdots s_2 s_1 t_{j-1} \cdots t_2 t_1\tau_{\sigma}^{-1})=\text{sgn }(r_m\cdots r_2 r_1\sigma^{-1})\\
&\Rightarrow& \text{sgn }(s_m \cdots s_2 s_1)\cdot \text{sgn }(t_{j-1} \cdots t_2 t_1)\cdot \text{sgn }(\tau_{\sigma}^{-1})=\text{sgn }(r_m\cdots r_2 r_1)\cdot \text{sgn }(\sigma^{-1})\\
&\Rightarrow& (-1)^{m}(-1)^{j-1}\text{sgn }(\tau_{\sigma}^{-1})=(-1)^{m}\text{sgn }(\sigma^{-1})\\
&\Rightarrow& (-1)^{j-1}\text{sgn }(\tau_{\sigma}^{-1})=\text{sgn }(\sigma^{-1})\\
&\Rightarrow& (-1)^{j-1}\text{sgn }(\tau_{\sigma})=\text{sgn }(\sigma)\\
&\Rightarrow& (-1)^{1+j}\text{sgn }(\tau_{\sigma})=\text{sgn }(\sigma)
\end{eqnarray*}
Best Answer
The wikiproof page in question uses a rather cryptic notation. The main point is the following: A partition of $\{1,...,n\}=H\sqcup H'$ into two subsets of (fixed) cardinalities $k$ and $n-k$, respectively, may be specified in a unique way through a permutation $r(1,...,n)=(r_1,...,r_k, r_{k+1},...r_n)$ where the ordered lists $r_1<\cdots < r_k$ and $r_{k+1} < \cdots <r_n$ run through the elements of the two subsets. Let $S_k(H) \times S_{n-k}(H')$ denote the set of permutations $\mu$ that shuffles elements within $H$ and within $H'$ but do not mix the two sets. Then every permutation $\rho\in S_n$ may be written in a unique way as $\rho=\mu\circ r$ for some $r$ and $\mu$ of the above-mentioned form.
An example with $n=5,k=2$ : Given the permutation $(1, 2, 3, 4, 5)\stackrel{\rho}\longrightarrow (5, 2; 4, 1, 3)$ (the semi-colon is just to underline the partitions) there is a unique splitting (with $H=\{2,5\}$, $H'=\{1,3,4\}$): $$(1, 2, 3, 4, 5) \stackrel{r}{\longrightarrow}(2, 5; 1, 3, 4) \stackrel{\mu}\longrightarrow (5, 2; 4, 1, 3). $$ In the equation you mention $\rho$ is any permutation but e.g. $\rho(r_1,...,r_k)$ is what I would prefer to call $\mu(r_1,...,r_k)$, since $\mu$ restricted to $\{r_1,...,r_k\}$ is indeed just a permutation of that set. Anyway, as $\mu$ is just the product of two permutations the signature of $\mu$ is the product of signatures (in the article writing $\rho$ instead of $\mu$), $$ {\rm sgn}(\mu(r_1,...,r_k)) \times {\rm sgn}(\mu (r_{k+1},...,r_n)).$$
For the signature of $r$ in my example you need $r_2-2= 5-2=3$ neighboring binary swaps to move five from position 2 to position 5 and then you need $r_1-1=1$ binary swap to move $2$ to position 2. Note that automatically this puts the remaining $n-k=3$ elements in the correct order! In general, the number of binary swaps is given by $$\sum_{i=1}^k (r_i-i)=\sum_{i=1}^k r_i - k(k+1)/2.$$ Thus, $${\rm sgn}(r)=(-1)^{\sum_{i=1} r_i} \times (-1)^{k(k+1)/2}.$$ Now use the same argument for the permutation of the second indices, $\sigma\circ s$. The last factor appears twice, but as $(-1)^{k(k+1)}=1$ it disappears and you obtain the formula you stated when you make the above interpretation of $\rho(r_1,...,r_k)$ etc. The sequel of the article makes sense when using this interpretation. Hope this helps.