[proofplan] We prove the identity by comparing two expansions of the Cauchy double alternant. One expansion expresses the determinant as a sum of products of alternants $a_\mu(x)a_\mu(y)$, while the other expansion factors out the Vandermonde alternants and records the remaining coefficients as Jacobi-Trudi determinants. Comparing the coefficient of $a_{\lambda+\delta}(y_1,\dots,y_n)$ gives $a_{\lambda+\delta}(x_1,\dots,x_n)=a_\delta(x_1,\dots,x_n)s_\lambda(x_1,\dots,x_n)$, and division by the nonzero alternant $a_\delta$ gives the stated quotient formula. [/proofplan] [step:Introduce the complete homogeneous symmetric functions and the Jacobi-Trudi form of $s_\lambda$] Let $R := \mathbb{Z}[x_1,\dots,x_n]$ be the [polynomial ring](/page/Polynomial%20Ring) in the variables $x_1,\dots,x_n$. Define \begin{align*} \delta := (n-1,n-2,\dots,0). \end{align*} For any $n$-tuple $\alpha=(\alpha_1,\dots,\alpha_n)$ of nonnegative integers and any variables $z_1,\dots,z_n$, define the alternant \begin{align*} a_\alpha(z_1,\dots,z_n) := \det\left(z_i^{\alpha_j}\right)_{1\leq i,j\leq n}. \end{align*} In particular, \begin{align*} a_\delta(z_1,\dots,z_n) = \prod_{1\leq i<j\leq n}(z_i-z_j). \end{align*} For each integer $r \in \mathbb{Z}$, define $h_r \in R$ by \begin{align*} h_r := \begin{cases} \displaystyle \sum_{\substack{\beta_1,\dots,\beta_n \geq 0\\ \beta_1+\cdots+\beta_n=r}} x_1^{\beta_1}\cdots x_n^{\beta_n}, & r \geq 0,\\ 0, & r < 0. \end{cases} \end{align*} Thus $h_0=1$, and $h_r$ is the complete homogeneous symmetric polynomial of degree $r$ in $x_1,\dots,x_n$ for $r \geq 0$. We use the [Jacobi-Trudi definition](/page/Jacobi-Trudi%20Identity) of the Schur polynomial: \begin{align*} s_\lambda(x_1,\dots,x_n) := \det\left(h_{\lambda_i-i+j}\right)_{1 \leq i,j \leq n}. \end{align*} It is therefore enough to prove \begin{align*} a_{\lambda+\delta}(x_1,\dots,x_n) = a_\delta(x_1,\dots,x_n) \det\left(h_{\lambda_i-i+j}\right)_{1 \leq i,j \leq n}. \end{align*} [/step] [step:Expand the Cauchy double alternant as a sum of alternants] Introduce a second set of indeterminates $y_1,\dots,y_n$, and work in the formal [power series](/page/Power%20Series) ring \begin{align*} \mathbb{Z}[x_1,\dots,x_n][[y_1,\dots,y_n]]. \end{align*} Consider the determinant \begin{align*} D(x,y) := \det\left(\frac{1}{1-x_i y_j}\right)_{1 \leq i,j \leq n}. \end{align*} Since \begin{align*} \frac{1}{1-x_i y_j} = \sum_{m=0}^{\infty} x_i^m y_j^m \end{align*} as a formal power series, multilinearity of the determinant in its columns gives \begin{align*} D(x,y) = \sum_{m_1,\dots,m_n \geq 0} \det\left(x_i^{m_j}\right)_{1 \leq i,j \leq n} \prod_{j=1}^n y_j^{m_j}. \end{align*} This equality is coefficientwise in $\mathbb{Z}[x_1,\dots,x_n][[y_1,\dots,y_n]]$: for a fixed monomial $y_1^{k_1}\cdots y_n^{k_n}$, only the tuple $(m_1,\dots,m_n)=(k_1,\dots,k_n)$ contributes. If two of the integers $m_1,\dots,m_n$ are equal, then the determinant $\det(x_i^{m_j})$ has two identical columns and is zero. Therefore only tuples with distinct entries contribute. Reordering each contributing tuple into strictly decreasing order $\mu_1>\cdots>\mu_n\geq 0$ gives one copy of the alternant in the $x$-variables and the same sign from reordering the $y$-monomial into the corresponding alternant. Hence \begin{align*} D(x,y) = \sum_{\mu_1>\cdots>\mu_n\geq 0} a_\mu(x_1,\dots,x_n)\,a_\mu(y_1,\dots,y_n), \end{align*} where $\mu=(\mu_1,\dots,\mu_n)$. [/step] [step:Factor the same determinant by the Cauchy double alternant identity] We now compute the same determinant as a rational function. Put all entries over the common denominator \begin{align*} P(x,y) := \prod_{i=1}^n\prod_{j=1}^n (1-x_i y_j). \end{align*} Then \begin{align*} P(x,y)D(x,y) \end{align*} is a polynomial which is alternating in the variables $x_1,\dots,x_n$ and also alternating in the variables $y_1,\dots,y_n$. Therefore it is divisible by both Vandermonde alternants \begin{align*} a_\delta(x_1,\dots,x_n) = \prod_{1\leq i<j\leq n}(x_i-x_j), \qquad a_\delta(y_1,\dots,y_n) = \prod_{1\leq i<j\leq n}(y_i-y_j). \end{align*} By the [Vandermonde divisibility property for alternating polynomials](/page/Alternating%20Polynomial), write \begin{align*} P(x,y)D(x,y) = a_\delta(x_1,\dots,x_n)a_\delta(y_1,\dots,y_n)Q(x,y) \end{align*} for the resulting polynomial quotient $Q \in \mathbb{Z}[x_1,\dots,x_n,y_1,\dots,y_n]$. Since the quotient of an alternating polynomial by the Vandermonde alternant is symmetric, $Q$ is symmetric in $x_1,\dots,x_n$ and also symmetric in $y_1,\dots,y_n$. We show that $Q$ is constant. For each fixed $i \in \{1,\dots,n\}$, every term of $P(x,y)D(x,y)$ has degree at most $n-1$ in $x_i$, because after clearing denominators the factor depending on $x_i$ is a product of exactly $n-1$ factors of the form $1-x_i y_j$. Use lexicographic monomial order in the $x$-variables with $x_1>x_2>\cdots>x_n$, treating coefficients as elements of $\mathbb{Z}[y_1,\dots,y_n]$. The leading monomial of $a_\delta(x_1,\dots,x_n)$ is $x_1^{n-1}x_2^{n-2}\cdots x_n^0$ with coefficient $1$. Since $Q$ is symmetric in the variables $x_1,\dots,x_n$, if $Q$ had positive degree in the $x$-variables, then its leading $x$-monomial would have the form $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ with $\alpha_1>0$. The leading monomial of the product $a_\delta(x_1,\dots,x_n)Q(x,y)$ would then be \begin{align*} x_1^{\alpha_1+n-1}x_2^{\alpha_2+n-2}\cdots x_n^{\alpha_n}, \end{align*} because leading monomials multiply in the polynomial ring over the integral domain $\mathbb{Z}[y_1,\dots,y_n]$. This monomial has $x_1$-degree greater than $n-1$, contradicting the individual-degree bound for $P(x,y)D(x,y)$. Thus $Q$ is independent of $x_1,\dots,x_n$. The same leading-monomial argument in the $y$-variables, using lexicographic order with $y_1>y_2>\cdots>y_n$ and treating coefficients as elements of $\mathbb{Z}[x_1,\dots,x_n]$, shows that $Q$ is independent of $y_1,\dots,y_n$. Hence $Q$ is an integer constant. To identify the constant, compare the coefficient of \begin{align*} M:=x_1^{n-1}x_2^{n-2}\cdots x_n^0 y_1^{n-1}y_2^{n-2}\cdots y_n^0. \end{align*} The coefficient of $M$ in $a_\delta(x_1,\dots,x_n)a_\delta(y_1,\dots,y_n)$ is $1$. In $P(x,y)D(x,y)$, the determinant expansion gives \begin{align*} P(x,y)D(x,y) = \sum_{\sigma \in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^n\prod_{j\neq \sigma(i)}(1-x_i y_j). \end{align*} To obtain $M$, one must choose $n-i$ factors from the $i$th row and $n-j$ factors involving $y_j$ from the $j$th column. The unique zero-one selection matrix with these row and column sums is the triangular matrix selecting exactly the pairs $(i,j)$ with $j\leq n-i$. This selection is compatible with the omitted factors only for the reverse permutation $\sigma(i)=n+1-i$. Its contribution has sign \begin{align*} \operatorname{sgn}(\sigma)(-1)^{n(n-1)/2} = (-1)^{n(n-1)/2}(-1)^{n(n-1)/2} = 1. \end{align*} Thus the coefficient of $M$ in $P(x,y)D(x,y)$ is also $1$, so the constant is $1$. Hence \begin{align*} D(x,y) = \frac{ a_\delta(x_1,\dots,x_n)a_\delta(y_1,\dots,y_n) }{ \prod_{i=1}^n\prod_{j=1}^n (1-x_i y_j) }. \end{align*} [guided] The determinant $D(x,y)$ has two different useful descriptions. The first description came from expanding each entry as a geometric series. The second description comes from treating the determinant as a rational function and clearing denominators. Define \begin{align*} P(x,y) := \prod_{i=1}^n\prod_{j=1}^n (1-x_i y_j). \end{align*} Then $P(x,y)D(x,y)$ is a polynomial. If we interchange two $x$-variables, say $x_i$ and $x_j$, then two rows of the determinant are interchanged, so the determinant changes sign. The factor $P(x,y)$ is symmetric in the $x$-variables, so $P(x,y)D(x,y)$ is alternating in $x_1,\dots,x_n$. The same argument with columns shows that it is alternating in $y_1,\dots,y_n$. Every alternating polynomial in $x_1,\dots,x_n$ is divisible by the Vandermonde alternant \begin{align*} a_\delta(x_1,\dots,x_n) = \prod_{1\leq i<j\leq n}(x_i-x_j), \end{align*} because setting $x_i=x_j$ forces the polynomial to vanish. Applying the same argument in the $y$-variables shows that \begin{align*} P(x,y)D(x,y) \end{align*} is divisible by \begin{align*} a_\delta(x_1,\dots,x_n)a_\delta(y_1,\dots,y_n). \end{align*} By the [Vandermonde divisibility property for alternating polynomials](/page/Alternating%20Polynomial), write \begin{align*} P(x,y)D(x,y) = a_\delta(x_1,\dots,x_n)a_\delta(y_1,\dots,y_n)Q(x,y) \end{align*} for the polynomial quotient $Q$. The quotient is symmetric in $x_1,\dots,x_n$ because both $P(x,y)D(x,y)$ and $a_\delta(x_1,\dots,x_n)$ change sign under every transposition of two $x$-variables, and the same argument gives symmetry in $y_1,\dots,y_n$. We must show that $Q$ has no remaining dependence on either set of variables. Fix $i \in \{1,\dots,n\}$. In each summand of the cleared determinant, the variable $x_i$ occurs in exactly the product of the $n-1$ factors $1-x_i y_j$ with one value of $j$ omitted. Hence $P(x,y)D(x,y)$ has degree at most $n-1$ in each individual variable $x_i$. Now $Q$ is symmetric in $x_1,\dots,x_n$, because both $P(x,y)D(x,y)$ and $a_\delta(x_1,\dots,x_n)$ are alternating in those variables. To avoid any possible cancellation issue, use lexicographic monomial order in the $x$-variables with $x_1>x_2>\cdots>x_n$, treating coefficients as elements of $\mathbb{Z}[y_1,\dots,y_n]$. The leading monomial of $a_\delta(x_1,\dots,x_n)$ is \begin{align*} x_1^{n-1}x_2^{n-2}\cdots x_n^0 \end{align*} with coefficient $1$. Suppose $Q$ had a nonconstant term involving the $x$-variables. Since $Q$ is symmetric, the leading $x$-monomial of $Q$ has the form $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ with $\alpha_1>0$. In a polynomial ring over an integral domain, the leading monomial of a product is the product of the leading monomials. Therefore the leading monomial of $a_\delta(x_1,\dots,x_n)Q(x,y)$ would have $x_1$-degree $\alpha_1+n-1>n-1$. This contradicts the fact that $P(x,y)D(x,y)$ has degree at most $n-1$ in $x_1$. Hence $Q$ is independent of the $x$-variables. Applying the corresponding lexicographic leading-monomial argument in the $y$-variables shows that $Q$ is independent of $y_1,\dots,y_n$. Therefore $Q$ is a constant. It remains to identify this constant. Compare the monomial \begin{align*} M:=x_1^{n-1}x_2^{n-2}\cdots x_n^0 y_1^{n-1}y_2^{n-2}\cdots y_n^0. \end{align*} The coefficient of $M$ in \begin{align*} a_\delta(x_1,\dots,x_n)a_\delta(y_1,\dots,y_n) \end{align*} is $1$. For $P(x,y)D(x,y)$, expanding the determinant after clearing denominators gives \begin{align*} P(x,y)D(x,y) = \sum_{\sigma \in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^n\prod_{j\neq \sigma(i)}(1-x_i y_j). \end{align*} To obtain $M$, the selected factors $x_i y_j$ must have row sums $n-i$ and column sums $n-j$. These row and column sums force the triangular selection $j\leq n-i$. This selection avoids exactly the omitted positions only for the reverse permutation $\sigma(i)=n+1-i$. Its sign contribution is \begin{align*} \operatorname{sgn}(\sigma)(-1)^{n(n-1)/2} = (-1)^{n(n-1)/2}(-1)^{n(n-1)/2} = 1. \end{align*} Thus the coefficient of $M$ in $P(x,y)D(x,y)$ is $1$, and the constant is $1$, giving \begin{align*} D(x,y) = \frac{ a_\delta(x_1,\dots,x_n)a_\delta(y_1,\dots,y_n) }{ \prod_{i=1}^n\prod_{j=1}^n (1-x_i y_j) }. \end{align*} [/guided] [/step] [step:Extract the coefficient of the alternant $a_{\lambda+\delta}(y)$] Let $t$ be a formal indeterminate. Using the generating function for the complete homogeneous symmetric polynomials in the formal power series ring $R[[t]]$, \begin{align*} \prod_{i=1}^n \frac{1}{1-x_i t} = \sum_{r=0}^{\infty} h_r t^r, \end{align*} we rewrite the factored expression as \begin{align*} D(x,y) &= a_\delta(x_1,\dots,x_n) a_\delta(y_1,\dots,y_n) \prod_{j=1}^n \left(\sum_{r=0}^{\infty} h_r y_j^r\right). \end{align*} Since \begin{align*} a_\delta(y_1,\dots,y_n) = \det\left(y_i^{n-j}\right)_{1\leq i,j\leq n}, \end{align*} multiplying the $i$th row by the series $\sum_{r=0}^{\infty}h_r y_i^r$ gives \begin{align*} a_\delta(y_1,\dots,y_n) \prod_{i=1}^n \left(\sum_{r=0}^{\infty} h_r y_i^r\right) = \det\left(\sum_{r=0}^{\infty} h_r y_i^{r+n-j}\right)_{1\leq i,j\leq n}. \end{align*} For each integer $k \geq 0$, the coefficient of $y_i^k$ in the $(i,j)$ entry is $h_{k-n+j}$, with the convention $h_r=0$ for $r<0$. Expanding the determinant coefficientwise is legitimate in the formal power series ring because, for each fixed total degree in $y_1,\dots,y_n$, only finitely many exponent choices can contribute. The coefficientwise alternating expansion gives \begin{align*} \det\left(\sum_{r=0}^{\infty} h_r y_i^{r+n-j}\right)_{1\leq i,j\leq n} = \sum_{\nu_1>\cdots>\nu_n\geq 0} \det\left(h_{\nu_i-n+j}\right)_{1\leq i,j\leq n} a_\nu(y_1,\dots,y_n). \end{align*} Indeed, the coefficient of the ordered monomial $y_1^{\nu_1}\cdots y_n^{\nu_n}$ is obtained by choosing from column $j$ of row $i$ the coefficient $h_{\nu_i-n+j}$, and summing over permutations of columns gives the determinant $\det(h_{\nu_i-n+j})_{1\leq i,j\leq n}$. Here $a_\nu(y_1,\dots,y_n)=\det(y_i^{\nu_j})_{1\leq i,j\leq n}$, so the ordering $\nu_1>\cdots>\nu_n$ fixes the alternant sign. Taking $\nu=\lambda+\delta$, so that \begin{align*} \nu_i=\lambda_i+n-i, \end{align*} the coefficient of $a_{\lambda+\delta}(y_1,\dots,y_n)$ in this expansion is \begin{align*} \det\left(h_{\lambda_i-i+j}\right)_{1\leq i,j\leq n} = s_\lambda(x_1,\dots,x_n). \end{align*} Therefore the coefficient of $a_{\lambda+\delta}(y_1,\dots,y_n)$ in the factored expansion of $D(x,y)$ is \begin{align*} a_\delta(x_1,\dots,x_n)s_\lambda(x_1,\dots,x_n). \end{align*} [/step] [step:Compare the two coefficients and divide by the Vandermonde alternant] From the alternant expansion of $D(x,y)$, the coefficient of $a_{\lambda+\delta}(y_1,\dots,y_n)$ is \begin{align*} a_{\lambda+\delta}(x_1,\dots,x_n). \end{align*} From the factored expansion of $D(x,y)$, the same coefficient is \begin{align*} a_\delta(x_1,\dots,x_n)s_\lambda(x_1,\dots,x_n). \end{align*} Equating the two coefficients gives the polynomial identity \begin{align*} a_{\lambda+\delta}(x_1,\dots,x_n) = a_\delta(x_1,\dots,x_n)s_\lambda(x_1,\dots,x_n). \end{align*} Since \begin{align*} a_\delta(x_1,\dots,x_n) = \prod_{1\leq i<j\leq n}(x_i-x_j) \end{align*} is a nonzero polynomial in the integral domain $\mathbb{Z}[x_1,\dots,x_n]$, we may divide by it in the fraction field $\mathbb{Q}(x_1,\dots,x_n)$. Hence \begin{align*} s_\lambda(x_1,\dots,x_n) = \frac{a_{\lambda+\delta}(x_1,\dots,x_n)}{a_\delta(x_1,\dots,x_n)}. \end{align*} This is the bialternant formula. [/step]