How to prove the cofactor formula for determinants, using a different definition of the determinant?

The proof in Wikipedia which you linked to in your previous attempt at asking this question is in fact using the very same definition of "determinant" that you are using, only with different terminology.

Wikipedia uses the definition of determinant as the sum of all terms of the form $$\mathrm{sgn}(\tau)a_{1\tau(1)}\cdots a_{n\tau(n)}$$ where $a_{ij}$ is the $(i,j)$th entry of the matrix, $\tau$ is a permutation of $\{1,2,\ldots,n\}$ (that is, a bijection $\tau\colon\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$), and $\mathrm{sgn}(\tau)$ is the sign of $\tau$.

In your description, the numbers $i_1,\ldots,i_n$ must be precisely the numbers $\{1,2,\ldots,n\}$ in some order; given a summand $$(-1)^ka_{1i_1}\cdots a_{ni_n}$$ as in your description, we define $\tau\colon\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$ by $\tau(j) = i_j$. Then $\tau$ is a permutation of $\{1,2,\ldots,n\}$, and we can express the summand as $$(-1)^ka_{1\tau(1)}\cdots a_{n\tau(n)}.$$ So the only thing left is to show that your factor $(-1)^k$ is precisely equal to $\mathrm{sgn}(\tau)$.

The sign of a permutation $\tau$ is equal to $1$ if $\tau$ can be written as a product of an even number of transpositions, and is equal to $-1$ if it can be written as a product of an odd number of transpositions. A "transposition" is precisely an "exchange of two elements".

For each "exchange" you perform to the list $i_1,i_2,\ldots,i_n$, consider the transposition that corresponds to that exchange. Performing the exchange is equivalent to composing $\tau$ with that transposition. The fact that after performing $k$ of these exchanges you get the identity tells you that $\tau$ can be written as a product of $k$ transpositions, and therefore that $\mathrm{sgn}(\tau)=(-1)^k$... exactly the same as the factor you have.

For instance, in your example $a_{13}a_{21}a_{34}a_{42}$, the permutation $\tau$ associated to this factor is, in disjoint cycle notation $\tau=(1,3,4,2)$, and in two-line notation $$\tau = \left(\begin{array}{cccc} 1&2&3&4\\ 3&1&4&2 \end{array}\right).$$ Exchanging the $1$ and $3$ is equivalent to composing $\tau$ with the transposition that exchanges $1$ and $3$ (which is $(1,3)$ in disjoint cycle notation); then exchanging $2$ and $3$; and finally exchanging $3$ and $4$, so you get $(-1)^3 = -1$. But this is the same as saying that $\tau$ is the composition of the three permutations I just mentioned: that is, that $(1,3,4,2) = (1,3)(2,3)(3,4)$, which is easy to verify. So the transposition has odd parity, hence $\mathrm{sgn}(\tau)=-1$, yielding the exact same result as yours.

Since the two definitions of determinant are in fact identical, except for the terminology used to describe the summands, the proof in Wikipedia is in fact "the proof [...] with [your] definition."