Interchanging two columns of a matrix will lead to change in the sign of determinant Interchanging two columns of a square matrix changes the sign of the determinant.
I know that this is true, and I do understand how it works. But is there any proof for this statement?

Any proof of this result depends on a definition of determinant. Let us define it using permutations: $\det(A) = \sum_{\tau \in S_n}\operatorname{sgn}(\tau)\,a_{1,\tau(1)}a_{2,\tau(2)} \ldots a_{n,\tau(n)},\;$ where the sum is over all $n!$ permutations of the columns by elements in the symmetric group $S_n.\;$ See the question about a determinant definition.
Let $A^\sigma$ be the result of rearranging the columns of $A$ using a permutation $\sigma.\;$ This replaces all the $\tau$ in the summation by $\sigma\tau$, the product of two permutations. Now $\;\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)\;$ and, by distributivity, the common $\operatorname{sgn}(\sigma)$ comes out of the summation. Thus, $\;\det(A^\sigma)=\operatorname{sgn}(\sigma)\det(A).$
In our case, interchanging any two columns is a transposition and these all have signature $-1$, and so multiplying the determinant by $-1$ changes its sign. QED.
