[proofplan] We prove the identity first as a polynomial identity in indeterminates $X_1,\dots,X_m$ over $F$, and then evaluate at $X_i = x_i$. The determinant polynomial is alternating in the variables, so every difference $X_i - X_j$ divides it. Since the product of all these differences has the same total degree as the determinant, the quotient is a constant; comparing one carefully chosen monomial shows that this constant is $1$. [/proofplan]

[step:Define the Vandermonde determinant polynomial over $F[X_1,\dots,X_m]$] Let $R := F[X_1,\dots,X_m]$ be the polynomial ring in $m$ indeterminates over $F$. Define the polynomial $P \in R$ by \begin{align*} P := \det \begin{pmatrix} 1 & 1 & \cdots & 1 \\ X_1 & X_2 & \cdots & X_m \\ \vdots & \vdots & & \vdots \\ X_1^{m-1} & X_2^{m-1} & \cdots & X_m^{m-1} \end{pmatrix}. \end{align*} Also define the polynomial $\Delta \in R$ by \begin{align*} \Delta := \prod_{1 \leq j < i \leq m} (X_i - X_j). \end{align*} We will prove $P = \Delta$ in $R$. [/step]

[step:Show that every factor $X_i - X_j$ divides the determinant polynomial]Fix indices $j,i \in \{1,\dots,m\}$ with $j < i$. View $P$ as a polynomial in the single indeterminate $X_i$ with coefficients in the ring $F[X_1,\dots,X_{i-1},X_{i+1},\dots,X_m]$. If we substitute $X_i = X_j$, then the $i$-th and $j$-th columns of the defining matrix for $P$ become equal, because for every row exponent $r \in \{0,\dots,m-1\}$ the two entries are both $X_j^r$. A determinant with two equal columns is zero, hence \begin{align*} P(X_1,\dots,X_{i-1},X_j,X_{i+1},\dots,X_m) = 0. \end{align*} By the factor theorem in the polynomial ring $F[X_1,\dots,X_{i-1},X_{i+1},\dots,X_m][X_i]$, the polynomial $X_i - X_j$ divides $P$ in $R$. Since this holds for every pair $1 \leq j < i \leq m$, and the linear polynomials $X_i - X_j$ are pairwise non-associate irreducible elements of the unique factorization domain $R$, their product $\Delta$ divides $P$. Thus there exists a polynomial $Q \in R$ such that \begin{align*} P = Q\Delta. \end{align*}[/step]

[guided]Fix a pair of indices $j < i$. The goal is to prove that $X_i - X_j$ is a factor of $P$. To do this, we isolate $X_i$ as the variable and treat all other $X_k$ as coefficients. Thus $P$ is an element of \begin{align*} F[X_1,\dots,X_{i-1},X_{i+1},\dots,X_m][X_i]. \end{align*} Now substitute $X_i = X_j$. In the Vandermonde matrix, the $i$-th column is \begin{align*} \begin{pmatrix} 1 \\ X_i \\ \vdots \\ X_i^{m-1} \end{pmatrix}, \end{align*} and the $j$-th column is \begin{align*} \begin{pmatrix} 1 \\ X_j \\ \vdots \\ X_j^{m-1} \end{pmatrix}. \end{align*} After the substitution $X_i = X_j$, these two columns are identical. A determinant with two equal columns is zero, so \begin{align*} P(X_1,\dots,X_{i-1},X_j,X_{i+1},\dots,X_m) = 0. \end{align*} The factor theorem now applies in the one-variable polynomial ring with coefficient ring $F[X_1,\dots,X_{i-1},X_{i+1},\dots,X_m]$: if a polynomial in $X_i$ vanishes at $X_i = X_j$, then it is divisible by $X_i - X_j$. Hence $X_i - X_j$ divides $P$. Doing this for every pair $j < i$ shows that each factor in \begin{align*} \Delta = \prod_{1 \leq j < i \leq m} (X_i - X_j) \end{align*} divides $P$. Because $F[X_1,\dots,X_m]$ is a unique factorization domain and these linear factors are irreducible and pairwise non-associate, the full product divides $P$. Therefore there is a polynomial $Q \in F[X_1,\dots,X_m]$ such that \begin{align*} P = Q\Delta. \end{align*}[/guided]

[step:Compare degrees to reduce the quotient to a constant] Every term in the determinant expansion of $P$ is a product containing exactly one entry from each row. The row exponents are $0,1,\dots,m-1$, so every nonzero term has total degree \begin{align*} 0 + 1 + \cdots + (m-1) = \frac{m(m-1)}{2}. \end{align*} Thus $P$ has total degree at most $\frac{m(m-1)}{2}$. The polynomial $\Delta$ is a product of exactly $\frac{m(m-1)}{2}$ linear factors, so $\Delta$ has total degree $\frac{m(m-1)}{2}$. Since $P = Q\Delta$, the polynomial $Q$ has total degree at most $0$. Hence $Q$ is a constant polynomial. Write this constant as $c \in F$, so \begin{align*} P = c\Delta. \end{align*} [/step]

[step:Compute the constant by comparing the monomial $X_1^0X_2^1\cdots X_m^{m-1}$] Let $M \in R$ denote the monomial \begin{align*} M := X_1^0 X_2^1 \cdots X_m^{m-1}. \end{align*} In $\Delta = \prod_{1 \leq j < i \leq m}(X_i - X_j)$, the monomial $M$ is obtained by choosing the term $X_i$ from every factor $X_i - X_j$. For fixed $i$, there are exactly $i-1$ factors of the form $X_i - X_j$ with $j < i$, so this choice contributes $X_i^{i-1}$. Its coefficient is therefore $1$. Now compute the coefficient of $M$ in $P$. By the Leibniz formula for the determinant, \begin{align*} P = \sum_{\sigma \in S_m} \operatorname{sgn}(\sigma) \prod_{r=1}^{m} X_{\sigma(r)}^{r-1}, \end{align*} where $S_m$ denotes the symmetric group on $\{1,\dots,m\}$ and $\operatorname{sgn}(\sigma) \in \{-1,1\}$ denotes the sign of $\sigma$. The monomial $M = X_1^0X_2^1\cdots X_m^{m-1}$ occurs only when $\sigma(r)=r$ for every $r \in \{1,\dots,m\}$, that is, only for the identity permutation. The sign of the identity permutation is $1$, so the coefficient of $M$ in $P$ is also $1$. Since $P = c\Delta$, comparing the coefficient of $M$ gives $1 = c \cdot 1$, hence $c=1$. Therefore \begin{align*} P = \Delta. \end{align*} [/step]

[step:Evaluate the polynomial identity at the given field elements] We now evaluate the identity $P=\Delta$ at the point $(x_1,\dots,x_m) \in F^m$. This gives \begin{align*} \det \begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_m \\ \vdots & \vdots & & \vdots \\ x_1^{m-1} & x_2^{m-1} & \cdots & x_m^{m-1} \end{pmatrix} = \prod_{1 \leq j < i \leq m} (x_i - x_j). \end{align*} This is exactly the claimed Vandermonde determinant formula. [/step]