[proofplan]
We expand each column of $A$ in the standard basis, use multilinearity to distribute, and apply the alternating property to eliminate all terms except those indexed by permutations of $\{1, \dots, n\}$. Each surviving term contributes $\varepsilon(\sigma) \cdot d(e_1, \dots, e_n)$ times the corresponding product of matrix entries, and the resulting sum is $(\det A) \cdot d(e_1, \dots, e_n)$.
[/proofplan]
[step:Expand columns in the standard basis and distribute by multilinearity]
Write $A^{(j)} = \sum_{i=1}^n A_{ij}\, e_i$ for each $j \in \{1, \dots, n\}$. By multilinearity of $d$:
\begin{align*}
d(A^{(1)}, \dots, A^{(n)}) = \sum_{i_1=1}^n \cdots \sum_{i_n=1}^n \left(\prod_{j=1}^n A_{i_j j}\right) d(e_{i_1}, \dots, e_{i_n}).
\end{align*}
[guided]
The key idea is that multilinearity lets us "distribute" the sum in each column. Since $A^{(j)} = \sum_{i_j=1}^n A_{i_j j}\, e_{i_j}$, multilinearity in the $j$th argument allows us to pull the sum and scalar outside. Doing this for all $n$ columns simultaneously produces $n^n$ terms, one for each choice of indices $(i_1, \dots, i_n) \in \{1, \dots, n\}^n$:
\begin{align*}
d(A^{(1)}, \dots, A^{(n)}) = \sum_{i_1=1}^n \cdots \sum_{i_n=1}^n \left(\prod_{j=1}^n A_{i_j j}\right) d(e_{i_1}, \dots, e_{i_n}).
\end{align*}
Most of these $n^n$ terms will vanish due to the alternating property -- only the $n!$ terms corresponding to permutations will survive.
[/guided]
[/step]
[step:Apply the alternating property to eliminate all non-permutation terms]
[claim:Alternating Kills Repeats]
If $i_k = i_\ell$ for some $k \neq \ell$, then $d(e_{i_1}, \dots, e_{i_n}) = 0$.
[/claim]
[proof]
The alternating property states $d(v_1, \dots, v_n) = 0$ whenever two arguments are equal. If $i_k = i_\ell$ with $k \neq \ell$, then the $k$th and $\ell$th arguments of $d$ are both $e_{i_k}$, so the value is zero.
[/proof]
Therefore the only non-zero terms have $(i_1, \dots, i_n)$ a permutation of $(1, \dots, n)$. Writing $i_j = \sigma(j)$ for $\sigma \in S_n$:
\begin{align*}
d(A^{(1)}, \dots, A^{(n)}) = \sum_{\sigma \in S_n} \left(\prod_{j=1}^n A_{\sigma(j),j}\right) d(e_{\sigma(1)}, \dots, e_{\sigma(n)}).
\end{align*}
[/step]
[step:Reduce $d(e_{\sigma(1)}, \dots, e_{\sigma(n)})$ to $\varepsilon(\sigma) \cdot d(e_1, \dots, e_n)$]
[claim:Sign From Transpositions]
For any $\sigma \in S_n$, $d(e_{\sigma(1)}, \dots, e_{\sigma(n)}) = \varepsilon(\sigma) \cdot d(e_1, \dots, e_n)$.
[/claim]
[proof]
The alternating property implies that swapping two adjacent arguments changes the sign. To see this, let $u$ and $v$ occupy positions $k$ and $k+1$. Evaluating $d(\dots, u + v, \dots, u + v, \dots) = 0$ (two equal arguments) and expanding by multilinearity:
\begin{align*}
d(\dots, u, \dots, u, \dots) + d(\dots, u, \dots, v, \dots) + d(\dots, v, \dots, u, \dots) + d(\dots, v, \dots, v, \dots) = 0.
\end{align*}
The first and fourth terms vanish (equal arguments), giving $d(\dots, u, \dots, v, \dots) = -d(\dots, v, \dots, u, \dots)$.
Since $\sigma$ can be written as a product of transpositions, and each transposition contributes a factor of $-1$, we obtain $d(e_{\sigma(1)}, \dots, e_{\sigma(n)}) = \varepsilon(\sigma) \cdot d(e_1, \dots, e_n)$.
[/proof]
[/step]
[step:Factor out $d(e_1, \dots, e_n)$ and identify $\det A$]
Substituting the claim into the sum:
\begin{align*}
d(A^{(1)}, \dots, A^{(n)}) &= \sum_{\sigma \in S_n} \varepsilon(\sigma) \left(\prod_{j=1}^n A_{\sigma(j),j}\right) d(e_1, \dots, e_n).
\end{align*}
Reindexing via $\tau = \sigma^{-1}$ (noting $\varepsilon(\sigma^{-1}) = \varepsilon(\sigma)$ and $\sigma \mapsto \sigma^{-1}$ is a bijection on $S_n$): setting $i = \sigma(j)$ so $j = \tau(i)$, the product $\prod_{j=1}^n A_{\sigma(j),j}$ becomes $\prod_{i=1}^n A_{i,\tau(i)}$. Therefore
\begin{align*}
d(A^{(1)}, \dots, A^{(n)}) = \left(\sum_{\tau \in S_n} \varepsilon(\tau) \prod_{i=1}^n A_{i,\tau(i)}\right) d(e_1, \dots, e_n) = (\det A) \cdot d(e_1, \dots, e_n).
\end{align*}
In particular, every alternating multilinear form is a scalar multiple of $\det$ (set the scalar to $d(e_1, \dots, e_n)$), so the space of such forms is at most one-dimensional, and $\det$ is the unique one normalised to $\det(e_1, \dots, e_n) = 1$.
[/step]
