Proof. We use induction. For \(n=1\) the condition \(A_{ij}=0\) for \(i>j\) is vacuous and (5.6) is trivially satisfied, thus the statement is anchored.

Inductive step: Assume \(n \in \mathbb{N}\) and \(n\geqslant 2.\) We want to show that if (5.6) holds for upper triangular \((n-1)\times(n-1)\)-matrices, then also for upper triangular \(n\times n\)-matrices. Let \(\mathbf{A}=(A_{ij})_{1\leqslant i,j\leqslant n} \in M_{n,n}(\mathbb{K})\) be an upper triangular matrix. Choosing \(l=1\) in the formula for \(\det(\mathbf{A}),\) we obtain \[\begin{aligned} \det(\mathbf{A})&=\sum_{k=1}^n(-1)^{k+1}A_{k1}\det\left(\mathbf{A}^{(k,1)}\right)=A_{11}\det\left(\mathbf{A}^{(1,1)}\right)+\sum_{k=2}^nA_{k1}\det\left(\mathbf{A}^{(k,1)}\right)\\ &=A_{11}\det\left(\mathbf{A}^{(1,1)}\right), \end{aligned}\] where the last equality uses that \(A_{k1}=0\) for \(k>1.\) We have \(\mathbf{A}^{(1,1)}=(A_{ij})_{2\leqslant i,j\leqslant n}\) and since \(\mathbf{A}\) is an upper triangular matrix, it follows that \(\mathbf{A}^{(1,1)}\) is an \((n-1)\times(n-1)\) upper triangular matrix as well. Hence by the induction hypothesis, we obtain \[\det(\mathbf{A}^{(1,1)})=\prod_{i=2}^nA_{ii}.\] We conclude that \(\det(\mathbf{A})=\prod_{i=1}^n A_{ii},\) as claimed.

We have \[\mathbf{P}_{\sigma \cdot \sigma^{-1}}=\mathbf{P}_{1}=\mathbf{1}_{n}=\mathbf{P}_{\sigma}\mathbf{P}_{\sigma^{-1}}\] showing that \(\mathbf{P}_{\sigma}\) is invertible with inverse \((\mathbf{P}_{\sigma})^{-1}=\mathbf{P}_{\sigma^{-1}}.\)

Inductive step: Assume \(n \in \mathbb{N}\) and \(n\geqslant 2.\) We want to show that if the claim holds for \(S_{n-1},\) then also for \(S_n.\) Let \(\sigma \in S_n\) and define \(k=\sigma(n).\) Then the permutation \(\sigma_1=\tau_{n,k}\sigma\) satisfies \(\sigma_1(n)=\tau_{n,k}\sigma(n)=\tau_{n,k}(k)=n\) and hence does not permute \(n.\) Restricting \(\sigma_1\) to the first \(n-1\) elements, we obtain a permutation of \(\{1,\ldots,n-1\}.\) By the induction hypothesis, we thus have \({\tilde{m}} \in \mathbb{N}\) and \(\tau_{k_1,l_1},\ldots \tau_{k_{{\tilde{m}}}},\tau_{l_{{\tilde{m}}}} \in S_n\) such that \[\sigma_1=\tau_{k_{{\tilde{m}}},l_{{\tilde{m}}}}\cdots \tau_{k_1,l_1}=\tau_{n,k}\sigma.\] Since \(\tau_{n,k}^2=1,\) multiplying from the left with \(\tau_{n,k}\) gives \(\sigma=\tau_{n,k}\tau_{k_{{\tilde{m}}},l_{{\tilde{m}}}}\cdots \tau_{k_1,l_1},\) the claim follows with \(m={\tilde{m}}+1.\)

Sketch of a proof. We can define a function \(f : \mathbb{K}^n \to \mathbb{K},\) \(\vec{x} \mapsto \det(\mathbf{V}_{\vec{x}}).\) By the Leibniz formula, the function \(f\) is a polynomial in the variables \(x_i\) with integer coefficients. If we freeze all variables of \(f\) except the \(\ell\)-th variable, then we obtain a function \(g_{\ell} : \mathbb{K}\to \mathbb{K}\) of one variable \(x_{\ell}.\) For \(1\leqslant i\leqslant n\) with \(i\neq \ell\) we have \(g_{\ell}(x_i)=0,\) since we compute the determinant of a matrix with two identical rows, the \(\ell\)-th row and the \(i\)-th row. Factoring the zeros, we can thus write \(g_{\ell}(x_{\ell})=q_{\ell}(x_{\ell})\prod_{1\leqslant i\leqslant n, i\neq \ell}(x_{\ell}-x_i)\) for some polynomial \(q_{\ell}.\) We can repeat this argument for all \(\ell\) and hence can write \(\det(\mathbf{V}_{\vec{x}})=q(\vec{x})\prod_{1\leqslant i<j\leqslant n}(x_j-x_i)\) for some polynomial \(q(\vec{x}).\) The Leibniz formula implies that the sum of the exponents of all the factors \(x_i\) in \(\det(\mathbf{V}_{\vec{x}})\) must be \(\frac{1}{2}n(n-1).\) The same holds true for \(\prod_{1\leqslant i<j \leqslant n}.\) It follows that \(q\) must be a constant. Using the Leibniz formula again, we see that the summand of \(\det(\mathbf{V}_{\vec{x}})\) corresponding to the identity permutation is the product of the diagonal entries of \(\mathbf{V}_{\vec{x}},\) that is, \(x_2(x_3)^2\cdots(x_n)^{n-1}.\) Taking the first term in all factors of \(\prod_{1\leqslant i<j\leqslant n}(x_j-x_i),\) we also obtain \(x_2(x_3)^2\cdots(x_n)^{n-1},\) hence \(\det(\mathbf{V}_{\vec{x}})=\prod_{1\leqslant i<j\leqslant n}(x_j-x_i),\) as claimed.