设 \(f:\mathcal{M}(\F) \to \F\)是满足行列式公理化定义中4个性质的映射。任取 \(n\)阶方阵 \(A\)，设
\begin{equation*}
A=\begin{pmatrix}
A_1\\
A_2\\
\vdots \\
A_n
\end{pmatrix},
\end{equation*}
其中 \(A_i\)是 \(A\)的第 \(i\)行，则
\begin{equation*}
A_i=\sum\limits_{j=1}^n a_{ij}\varepsilon_j^T,
\end{equation*}
这里\(\varepsilon_j^T\)表示第 \(j\)个元素是 \(1\)的 \(n\)维单位行向量。根据行列式公理化定义的最后两条性质，得
\begin{equation}
\begin{array}{rl}
f(A)= & \sum\limits_{j_1=1}^n a_{1j_1} f\left(\begin{array}{c}
\varepsilon_{j_1}^T\\
\sum\limits_{j=1}^n a_{2j}\varepsilon_j^T\\
\vdots \\
\sum\limits_{j=1}^n a_{nj}\varepsilon_j^T
\end{array}\right)\\
= & \sum\limits_{j_1=1}^n\sum\limits_{j_2=1}^n a_{1j_1}a_{2j_2}f\left(\begin{array}{c}
\varepsilon_{j_1}^T\\
\varepsilon_{j_2}^T\\
\vdots \\
\sum\limits_{j=1}^n a_{nj}\varepsilon_j^T
\end{array}\right)\\
= & \cdots \\
= & \sum\limits_{j_1,j_2,\ldots ,j_n=1}^n a_{1j_1}a_{2j_2}\cdots a_{nj_n}f\left(\begin{array}{c}
\varepsilon_{j_1}^T\\
\varepsilon_{j_2}^T\\
\vdots \\
\varepsilon_{j_n}^T
\end{array}\right).
\end{array}\tag{3.2.2}
\end{equation}
由公理化定义的第\(2\)条性质不难验证，当\(j_i=j_k\)时，
\begin{equation*}
f\left(\begin{array}{c}
\varepsilon_{j_1}^T\\
\varepsilon_{j_2}^T\\
\vdots \\
\varepsilon_{j_n}^T
\end{array}\right)=0.
\end{equation*}
因此在
(3.2.2)中，只剩下
\(j_1,j_2,\ldots ,j_n\)互不相同的项，则
\begin{equation*}
f(A)=\sum\limits_{j_1,j_2,\ldots ,j_n} a_{1j_1}a_{2j_2}\cdots a_{nj_n}f\left(\begin{array}{c}
\varepsilon_{j_1}^T\\
\varepsilon_{j_2}^T\\
\vdots \\
\varepsilon_{j_n}^T
\end{array}\right).
\end{equation*}
这里\(j_1,j_2,\ldots ,j_n\)取遍 \(1,2,\ldots ,n\)的所有排列。因为\(j_1,j_2,\ldots ,j_n\)可经过 \(\tau(j_1,j_2,\ldots ,j_n)\)次对换变成 \(1,2,\ldots ,n\)，所以根据公理化定义的前两条性质
\begin{align*}
f(A)&=\sum\limits_{j_1,j_2,\ldots ,j_n} (-1)^{\tau(j_1,j_2,\ldots ,j_n)}a_{1j_1}a_{2j_2}\cdots a_{nj_n}f\left(\begin{array}{c}
\varepsilon_{1}^T\\
\varepsilon_{2}^T\\
\vdots \\
\varepsilon_{n}^T
\end{array}\right) \\
& =\sum\limits_{j_1,j_2,\ldots ,j_n} (-1)^{\tau(j_1,j_2,\ldots ,j_n)}a_{1j_1}a_{2j_2}\cdots a_{nj_n}.
\end{align*}
