The determinant is alternating

1. Equal columns. Suppose columns $j$ and $k$ (with $j\neq k$) are equal. Consider the transposition $\tau=(j\,k)$ in $S_n$. For each permutation $\sigma$, $\sigma\circ\tau$ has the same sign as $\sigma$ (since $\varepsilon(\tau)=-1$ and $\varepsilon(\sigma\circ\tau)=-\varepsilon(\sigma)$). Using the Leibniz formula, pair up the terms:
$$\det A = \sum_\sigma \varepsilon(\sigma)\prod_i a_{\sigma(i),i} = \sum_\sigma \varepsilon(\sigma\circ\tau)\prod_i a_{\sigma\circ\tau(i),i}.$$
But because columns $j$ and $k$ are equal, $\prod_i a_{\sigma\circ\tau(i),i} = \prod_i a_{\sigma(i),i}$. So the two sums cancel, giving $\det A = -\det A$, hence $\det A=0$.

2. Swapping columns. Let $B$ be obtained from $A$ by swapping columns $j$ and $k$. Consider the determinant of $A+B$: column $j$ of $A+B$ is $C_j + C_k$, and column $k$ is $C_k+C_j$, so these columns are equal. By (1), $\det(A+B)=0$. By multilinearity:
$$0 = \det(A+B) = \det A + \det A^{j\leftarrow k} + \det A^{k\leftarrow j} + \det B,$$
where $A^{j\leftarrow k}$ has column $j$ replaced by $C_k$ (and column $k$ unchanged). This matrix has two equal columns $C_k$, so its determinant is zero. Similarly $\det A^{k\leftarrow j}=0$. Hence $\det B = -\det A$.

Conclusion. The determinant is alternating: zero on equal columns, and swapping two columns flips the sign.