1. By interchanging the two arguments that are equal to the vector #\vec{a}#, the determinant transforms into its negative, but at the same time, the arguments do not change, so the determinant remains the same. This is only possible if the determinant is #0#.

2. Assume for the sake of convenience that \[ \vec{a}_1=\alpha_2\vec{a}_2+\cdots +\alpha_n\vec{a}_n \] Then \[\begin{array}{rl}D(\vec{a}_1,\vec{a}_2,\ldots ,\vec{a}_n)& =D(\sum_{k=2}^n\alpha_k\vec{a}_k,\vec{a}_2,\ldots ,\vec{a}_n)\\& =\sum_{k=2}^n\alpha_k D(\vec{a}_k ,\vec{a}_2,\ldots ,\vec{a}_n)\\&=0\end{array}\] because of the first part.

Uniqueness: We are now able to write down the determinantal function for each #n#. It is done in the same way as for #(2\times 2)#-matrices: we will write each row as a linear combination of the standard basis vectors and use the multilinearity to write the determinant as a sum of many determinants of standard basis vectors. For this purpose consider the matrix
\[
A=\left(\,\begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1n}\\
a_{21} & a_{22} & \ldots & a_{2n}\\
\vdots & \vdots & \vdots & \vdots\\
a_{n1} & \ldots & \ldots & a_{nn}
\end{array}\,\right)
\] with rows #\vec{a}_1 ,\ldots ,\vec{a}_n#. Then, for each #i#,
\[
\vec{a}_i=\sum_{j=1}^n\ a_{ij}\vec{e}_j
\] and so\[
\begin{array}{ll}
D\ (\vec{a}_1 ,\ldots ,\vec{a}_n)\
&=\ D\Big(\sum_{j_1=1}^na_{1j_1}\vec{e}_{j_1},\ldots ,\sum_{j_n=1}^n a_{nj_n}\vec{e}_{j_n}\Big)\\
&=\ \sum_{j_1=1}^n\ldots \sum_{j_n=1}^na_{1j_1}\ldots a_{nj_n}D(\vec{e}_{j_1},\ldots ,\vec{e}_{j_n})
\end{array}
\]This is a sum of a lot of terms. There are #n# summation indices, each with #n# possible values, so the number of terms is #n^n#. For a value as low as #n=8# we have \(16\,777\,216\) terms.

Since terms in which two indices #j_i# and #j_k# are equal constribute zero, we we can limit ourselves to permutations \(\rv{j_1,\ldots ,j_n}\) of #\{1,2,\ldots,n\}#. We count these permutations as follows. For the first element, there are #n# possibilities, then for the second element there are #n-1# possibilities, for the third element #n-2#, #\ldots#, for the penultimate #2#, and finally, the last element is fixed. Thus, there are #n\cdot(n-1)\cdots 2\cdot 1=n!# permutations of #\{1,\ldots ,n\}#.

The sum for #D# thus has #n!# terms. That is a lot less than #n^n#, but for #n=8# there still are #8!=40\,320# terms. If some terms #D(\vec{e}_{j_1},\ldots ,\vec{e}_{j_n})# of the sum over all permutations \(\rv{j_1,\ldots ,j_n}\) would be equal to #0#, then the sum would consist of fewer terms. However, this is not the case. Because #\rv{j_1,\ldots, j_n}# contains all numbers #1,\ldots ,n#, we can obtain the order #1,\ldots ,n# by repeated transpositions. Each transposition involves a factor #-1#. If the number of permutations is even, then #D(\vec{e}_{j_1},\ldots ,\vec{e}_{j_n})# equals #1# and otherwise it is equal to #-1#. Thus, we find:

If #D# exists, then
\[
D(\vec{a}_1,\ldots ,\vec{a}_n)=\sum_{j_1,\ldots ,j_n}\pm a_{1j_1}\ldots a_{nj_n}
\] where the sum runs over all #n!# permutations of #\rv{1,\ldots ,n}# and the sign must be #1# if the permutation of the sequence #\rv{1,\ldots ,n}# involves an even number of transpositions and #-1# otherwise.

det is a determinantal function: To prove this we verify that #\det# meets the three requirements for a determinantal function.

Multilinearity: Let #i# be one of the numbers #1,\ldots,n#. We want to show that #\det# is linear in the #i#-th argument #\vec{a}_i#. Because \( \det(\vec{a}_1,\ldots ,\vec{a}_n)\) is a sum of terms of the form \(\text{sg}(\sigma) a_{1\sigma(1)}\cdots a_{n\sigma(n)}\), it suffices to verify that each of these terms is linear in #\vec{a}_i#. That is indeed the case, because only the factor #a_{i\sigma(i)}# comes from #\vec{a}_i#.

Antisymmetry: upon an interchange of two vectors in the arguments, the value of the determinant changes to its negative. If we interchange the arguments #i# and #j#, we get the same expression, with #a_{i\sigma(i)}# and #a_{j\sigma(j)}# in each term replaced by #a_{j\sigma(i)}# and #a_{i\sigma(j)}#, respectively. Because it only involves a transposition, we may suppose #i\lt j#. So if we let #\tau# be the composition of #\sigma# and the transposition of #i# and #j#, we get

\[\begin{array}{rcl}&&a_{1\sigma(1)}\cdots a_{i-1\sigma(i-1)}a_{j\sigma(i)}a_{i+1\sigma(i+1)}\cdots a_{j-1\sigma(j-1)}a_{i\sigma(j)}a_{j+1\sigma(j+1)}\cdots a_{n\sigma(n)}\\&&\phantom{xx}=a_{1\tau(1)}\cdots a_{i-1\tau(i-1)}a_{j\tau(j)}a_{i+1\tau(i+1)}\cdots a_{j-1\tau(j-1)}a_{i\tau(i)}a_{j+1\tau(j+1)}\cdots a_{n\tau(n)}\\ &&\phantom{xx}= a_{1\tau(1)}\cdots a_{n\tau(n)}\end{array}\] For clarity, we treat these equalities again, but with more words:

\[\begin{array}{lcc}&{\left.a_{1\sigma(1)}\cdots a_{n\sigma(n)}\right.}&\qquad\left({\text{with }a_{i\sigma(i)} \text{ and }a_{j\sigma(j)}\text{ replaced by }a_{j\sigma(i)}\text{ and }a_{i\sigma(j)} }\right)\\ &\phantom{1234}={\left.a_{1\tau(1)}\cdots a_{n\tau(n)}\right.}&\qquad\left({\text{with }a_{i\tau(i)}\text{ and }a_{j\tau(j)}\text{ interchanged} }\right)\\ &\phantom{1234}=a_{1\tau(1)}\cdots a_{n\tau(n)}&\qquad\left(\text{in a product of scalars the sequence does not matter}\right)\end{array}\]

From #\tau\equiv\sigma\,(i,j)# it follows that #\sigma# has one transposition less than #\tau# such that #\text{sg}(\sigma)=-\text{sg}(\tau)#. The sum over all permutations #\sigma# equals the sum over all permutations #\tau#. With all these results we find:

\[\begin{array}{rcll}\det\overbrace{(\vec{a}_1,\ldots ,\vec{a}_n)}^{\vec{a}_i\quad\leftrightarrow\quad\vec{a}_j}&=&\sum_{\sigma}\text{sg}(\sigma)\cdot \overbrace{a_{1\sigma(1)}\cdots a_{n\sigma(n)}}^{\begin{array}{rcl}a_{i\sigma(i)}&\leftrightarrow&a_{j\sigma(i)}\\a_{j\sigma(j)}&\leftrightarrow&a_{i\sigma(j)}\end{array}}&\color{blue}{\text{definition }\det}\\&=&\sum_{\sigma}\text{sg}(\sigma)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\text{result from above}}\\&=&\sum_{\sigma}-\text{sg}(\tau)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\text{sg}(\sigma)=-\text{sg}(\tau)}\\&=&\sum_{\tau}-\text{sg}(\tau)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\sum_{\sigma}=\sum_{\tau}}\\&=&-\sum_{\tau}\text{sg}(\tau)\cdot a_{1\tau(1)}\cdots a_{n\tau(n)}&\color{blue}{\text{minus sign moved forward}}\\&=&-\det(\vec{a}_1,\ldots ,\vec{a}_n)&\color{blue}{\text{definition }\det}\end{array}\]

Normalization: #D(\vec{e}_1,\vec{e}_2,\ldots ,\vec{e}_n)=1#.

A standard basis vector #\vec{e}_i# of #\mathbb{R}^n# consists of #n-1# zeroes and one #1# at position #i#:\[\vec{e}_i=\rv{e_{i1},e_{i2},\ldots,e_{in}}=[0,0,\ldots,0,\underbrace{1}_{\text{position }i},0,\ldots,0,0]\]This means:\[e_{ij}=\left\{\begin{array}{rl}1&\text{for }i=j\\0&\text{for }i\neq j\end{array}\right.\]Below, in the sum over all permutations #\sigma# the only term unequal to zero is therefore #e_{11}\cdots e_{nn}# belonging to #\sigma=\rv{1,2,\ldots,n}#:\[\begin{array}{rcl}\det(\vec{e}_1,\vec{e}_2,\ldots ,\vec{e}_n) &=&\sum_{\sigma}\text{sg}(\sigma)\cdot e_{1\sigma(1)}\cdots e_{n\sigma(n)}\\ &=&\text{sg}(\rv{1,\ldots,n})\cdot e_{11}\cdots e_{nn}\\&=& 1\cdot 1\cdots 1\\&=& 1\end{array}\]

Last statement: If #E(I) = 0#, then each term in the expansion of #E(A)# is zero because it contains a factor #E(I)#. Then #E(A) = 0 = E(I)\cdot \det(A)#, as required. Suppose, therefore, #E(I)\ne0#. Then #\frac{E(A)}{E(I)}# is defined; it is a function that satisfies all three requirements for a determinantal function so because of the uniqueness we have \[ \frac{E(A)}{E(I)} = \det(A)\] The statement now follows after multiplying both members by #E(I)#.