[proofplan] We prove the identity first in the [polynomial ring](/page/Polynomial%20Ring) $\mathbb{Z}[x_1,\dots,x_N]$ for an arbitrary finite alphabet, and then pass to symmetric functions by stability in $N$. The key input is the generating series for the complete homogeneous symmetric polynomials, which converts the Jacobi-Trudi determinant into a quotient of two alternants. Multiplying by the Vandermonde determinant reduces the identity to an explicit coefficient extraction from a determinant product. The resulting alternant is exactly the bialternant formula for the Schur polynomial, so the determinant equals $s_\lambda$. [/proofplan]

[step:Work in a finite alphabet and record the complete homogeneous generating series]Fix an integer $N \geq r$. Let \begin{align*} x &:=(x_1,\dots,x_N) \end{align*} be an $N$-variable alphabet, and for each integer $m$ define $h_m(x) \in \mathbb{Z}[x_1,\dots,x_N]$ by \begin{align*} h_m(x) &= \begin{cases} \displaystyle \sum_{1 \leq a_1 \leq \cdots \leq a_m \leq N} x_{a_1}\cdots x_{a_m}, & m \geq 1,\\ 1, & m=0,\\ 0, & m<0. \end{cases} \end{align*} Equivalently, the formal [power series](/page/Power%20Series) \begin{align*} H_x(t): \mathbb{Z}[[t]] &\to \mathbb{Z}[x_1,\dots,x_N][[t]] \end{align*} given by \begin{align*} H_x(t) := \sum_{m=0}^{\infty} h_m(x)t^m \end{align*} satisfies \begin{align*} H_x(t)=\prod_{a=1}^{N}(1-x_a t)^{-1}. \end{align*} This follows by expanding each factor as the geometric series \begin{align*} (1-x_a t)^{-1}=\sum_{q=0}^{\infty}x_a^q t^q \end{align*} and collecting the coefficient of $t^m$.[/step]

[guided]We first avoid any ambiguity about infinite symmetric functions by proving a finite-variable statement. Fix $N \geq r$ and work in $\mathbb{Z}[x_1,\dots,x_N]$. For $m \geq 1$, the polynomial $h_m(x)$ is the generating function for weakly increasing words of length $m$ in the alphabet $\{1,\dots,N\}$: \begin{align*} h_m(x)=\sum_{1 \leq a_1 \leq \cdots \leq a_m \leq N} x_{a_1}\cdots x_{a_m}. \end{align*} The conventions $h_0(x)=1$ and $h_m(x)=0$ for $m<0$ are part of the definition. The generating series packages these polynomials in a way that will make the determinant computable: \begin{align*} H_x(t):=\sum_{m=0}^{\infty} h_m(x)t^m. \end{align*} Expanding each geometric series gives \begin{align*} \prod_{a=1}^{N}(1-x_a t)^{-1} = \prod_{a=1}^{N}\left(\sum_{q=0}^{\infty}x_a^q t^q\right). \end{align*} The coefficient of $t^m$ in this product is obtained by choosing nonnegative integers $q_1,\dots,q_N$ with $q_1+\cdots+q_N=m$, and contributes the monomial $x_1^{q_1}\cdots x_N^{q_N}$. This is exactly the same data as a weakly increasing word of length $m$: the integer $q_a$ records how many times the letter $a$ appears. Therefore \begin{align*} H_x(t)=\prod_{a=1}^{N}(1-x_a t)^{-1}. \end{align*}[/guided]

[step:Express the Jacobi-Trudi determinant through coefficient extraction]Define the finite Jacobi-Trudi polynomial-valued map \begin{align*} J_{\lambda,N}: \mathbb{Z}^N &\to \mathbb{Z} \\ x &\mapsto \det\bigl(h_{\lambda_i-i+j}(x)\bigr)_{1 \leq i,j \leq r}. \end{align*} Equivalently, $J_{\lambda,N}(x)$ denotes this determinant as an element of $\mathbb{Z}[x_1,\dots,x_N]$. For each $1 \leq i \leq r$, define the integer \begin{align*} \alpha_i:=\lambda_i-i+r. \end{align*} Since $\lambda$ is a partition, the integers $\alpha_1,\dots,\alpha_r$ are pairwise distinct. For variables $t_1,\dots,t_r$, define the alternant polynomial-valued map \begin{align*} A_{\delta_r}: \mathbb{Z}^r &\to \mathbb{Z} \\ t &\mapsto \det\bigl(t_i^{r-j}\bigr)_{1 \leq i,j \leq r} =\prod_{1 \leq i<j \leq r}(t_i-t_j), \end{align*} where $\delta_r:=(r-1,r-2,\dots,0)$ and $t=(t_1,\dots,t_r)$. We claim that \begin{align*} J_{\lambda,N}(x) = [t_1^{\alpha_1}\cdots t_r^{\alpha_r}] \left( \prod_{i=1}^{r}H_x(t_i)\,A_{\delta_r}(t) \right), \end{align*} where $[t_1^{\alpha_1}\cdots t_r^{\alpha_r}]$ denotes coefficient extraction. Let $S_r$ denote the symmetric group on $\{1,\dots,r\}$, and let $\operatorname{sgn}:S_r\to\{ -1,1\}$ denote the [sign homomorphism](/theorems/778). Expanding the determinant defining $A_{\delta_r}(t)$ by permutations gives \begin{align*} A_{\delta_r}(t) = \sum_{\sigma \in S_r}\operatorname{sgn}(\sigma) \prod_{i=1}^{r}t_i^{r-\sigma(i)}. \end{align*} Thus \begin{align*} \prod_{i=1}^{r}H_x(t_i)A_{\delta_r}(t) = \sum_{\sigma \in S_r}\operatorname{sgn}(\sigma) \prod_{i=1}^{r} \left(\sum_{m_i=0}^{\infty}h_{m_i}(x)t_i^{m_i+r-\sigma(i)}\right). \end{align*} The coefficient of $t_i^{\alpha_i}=t_i^{\lambda_i-i+r}$ is obtained precisely when \begin{align*} m_i+r-\sigma(i)=\lambda_i-i+r, \end{align*} equivalently \begin{align*} m_i=\lambda_i-i+\sigma(i). \end{align*} Since $h_m(x)=0$ for $m<0$, the extracted coefficient is \begin{align*} \sum_{\sigma \in S_r}\operatorname{sgn}(\sigma) \prod_{i=1}^{r}h_{\lambda_i-i+\sigma(i)}(x), \end{align*} which is exactly the permutation expansion of $J_{\lambda,N}(x)$.[/step]

[guided]The point of introducing $A_{\delta_r}(t)$ is that its permutation expansion supplies exactly the column index in the Jacobi-Trudi determinant. We define the polynomial-valued map \begin{align*} A_{\delta_r}: \mathbb{Z}^r &\to \mathbb{Z} \\ t &\mapsto \det\bigl(t_i^{r-j}\bigr)_{1 \leq i,j \leq r}, \end{align*} where $\delta_r=(r-1,r-2,\dots,0)$ and $t=(t_1,\dots,t_r)$. Let $S_r$ denote the symmetric group on $\{1,\dots,r\}$, and let $\operatorname{sgn}:S_r\to\{ -1,1\}$ denote the sign homomorphism. Expanding this determinant along permutations gives \begin{align*} A_{\delta_r}(t) = \sum_{\sigma \in S_r}\operatorname{sgn}(\sigma) \prod_{i=1}^{r}t_i^{r-\sigma(i)}. \end{align*} This is the place where the earlier product $\prod_{i<j}(1-t_j/t_i)$ must be handled with care: the denominator needed to convert $\det(t_i^{r-j})$ into that product is $\prod_i t_i^{r-i}$, not $\prod_i t_i^{r-1}$. To avoid that bookkeeping error, we use the polynomial alternant $A_{\delta_r}(t)$ directly. Now multiply by the complete homogeneous generating functions. Since \begin{align*} H_x(t_i)=\sum_{m_i=0}^{\infty}h_{m_i}(x)t_i^{m_i}, \end{align*} we have \begin{align*} \prod_{i=1}^{r}H_x(t_i)A_{\delta_r}(t) = \sum_{\sigma \in S_r}\operatorname{sgn}(\sigma) \prod_{i=1}^{r} \left(\sum_{m_i=0}^{\infty}h_{m_i}(x)t_i^{m_i+r-\sigma(i)}\right). \end{align*} To extract the coefficient of $t_1^{\alpha_1}\cdots t_r^{\alpha_r}$, with $\alpha_i=\lambda_i-i+r$, we match exponents variable by variable. For a fixed permutation $\sigma$, the exponent of $t_i$ is $m_i+r-\sigma(i)$, so it equals $\alpha_i$ exactly when \begin{align*} m_i+r-\sigma(i)=\lambda_i-i+r. \end{align*} Solving this equation gives \begin{align*} m_i=\lambda_i-i+\sigma(i). \end{align*} If this integer is negative, the corresponding summand contributes $0$ because $h_m(x)=0$ for $m<0$. Therefore the coefficient extraction gives \begin{align*} [t_1^{\alpha_1}\cdots t_r^{\alpha_r}] \left( \prod_{i=1}^{r}H_x(t_i)A_{\delta_r}(t) \right) = \sum_{\sigma \in S_r}\operatorname{sgn}(\sigma) \prod_{i=1}^{r}h_{\lambda_i-i+\sigma(i)}(x), \end{align*} which is the permutation expansion of \begin{align*} \det\bigl(h_{\lambda_i-i+j}(x)\bigr)_{1 \leq i,j \leq r}. \end{align*}[/guided]

[step:Convert the coefficient formula into the finite Schur polynomial]Extend $\lambda$ to a sequence of length $N$ by setting $\lambda_b:=0$ for $r<b\leq N$, and define \begin{align*} \delta_N:=(N-1,N-2,\dots,0). \end{align*} Define the $N$ by $N$ Jacobi-Trudi polynomial-valued map \begin{align*} D_{\lambda,N}: \mathbb{Z}^N &\to \mathbb{Z} \\ x &\mapsto \det\bigl(h_{\lambda_i-i+j}(x)\bigr)_{1 \leq i,j \leq N}. \end{align*} Equivalently, $D_{\lambda,N}(x)$ denotes this determinant as an element of $\mathbb{Z}[x_1,\dots,x_N]$. For $i>r$ and $j<i$, one has $\lambda_i-i+j=j-i<0$, while for $i=j$ one has $\lambda_i-i+j=0$. Hence the matrix defining $D_{\lambda,N}(x)$ is block upper triangular with lower-right diagonal entries equal to $1$, and therefore \begin{align*} D_{\lambda,N}(x)=J_{\lambda,N}(x). \end{align*} Applying the coefficient formula from the preceding step with $r$ replaced by $N$ gives \begin{align*} D_{\lambda,N}(x) = [t_1^{\lambda_1+N-1}t_2^{\lambda_2+N-2}\cdots t_N^{\lambda_N}] \left( \prod_{i=1}^{N}H_x(t_i)A_{\delta_N}(t) \right). \end{align*} Write $[t^{\lambda+\delta_N}]$ for this coefficient extraction. Using $H_x(t_i)=\prod_{a=1}^{N}(1-x_a t_i)^{-1}$, this is \begin{align*} D_{\lambda,N}(x) = [t^{\lambda+\delta_N}] \left( A_{\delta_N}(t)\prod_{i=1}^{N}\prod_{a=1}^{N}(1-x_a t_i)^{-1} \right). \end{align*} In the finite-variable setting we use the standard characterization of the Schur polynomial by the [bialternant formula](/page/Schur%20Polynomial), equivalently to the tableau definition of $s_\lambda(x)$: \begin{align*} s_\lambda(x):=\frac{A_{\lambda+\delta_N}(x)}{A_{\delta_N}(x)}, \end{align*} where \begin{align*} A_{\lambda+\delta_N}(x)&:=\det\bigl(x_a^{\lambda_b+N-b}\bigr)_{1 \leq a,b \leq N},\\ A_{\delta_N}(x)&:=\det\bigl(x_a^{N-b}\bigr)_{1 \leq a,b \leq N} =\prod_{1 \leq a<b \leq N}(x_a-x_b). \end{align*} We now compute the coefficient directly. Define the formal power series-valued map \begin{align*} K: \mathbb{Z}^N \times \mathbb{Z}^N &\to \mathbb{Z}[[t_1,\dots,t_N]] \\ (x,t) &\mapsto A_{\delta_N}(t)\prod_{i=1}^{N}\prod_{a=1}^{N}(1-x_a t_i)^{-1}, \end{align*} where $t=(t_1,\dots,t_N)$. The following determinant identity gives \begin{align*} A_{\delta_N}(x)K(x,t) = \det\left((1-x_a t_i)^{-1}\right)_{1 \leq a,i \leq N}. \end{align*} Indeed, after multiplying the determinant on the right by $\prod_{a=1}^{N}\prod_{i=1}^{N}(1-x_a t_i)$, the result is alternating in the $x$ variables and in the $t$ variables. In each variable $x_a$ and in each variable $t_i$, its degree is at most $N-1$. Hence divisibility by $A_{\delta_N}(x)A_{\delta_N}(t)$ forces the quotient to have degree $0$ in every variable, so the quotient is constant. The coefficient of $x_1^{N-1}x_2^{N-2}\cdots x_N^0t_1^{N-1}t_2^{N-2}\cdots t_N^0$ is $1$ on both sides, so this constant is $1$. Expanding the determinant by multilinearity in its columns and using the geometric series in the formal power series ring gives \begin{align*} \det\left((1-x_a t_i)^{-1}\right)_{1 \leq a,i \leq N} = \sum_{k_1,\dots,k_N \geq 0} \det\bigl(x_a^{k_i}\bigr)_{1 \leq a,i \leq N} \prod_{i=1}^{N}t_i^{k_i}. \end{align*} Therefore the coefficient of $t^{\lambda+\delta_N}$ in $A_{\delta_N}(x)K(x,t)$ is \begin{align*} \det\bigl(x_a^{\lambda_i+N-i}\bigr)_{1 \leq a,i \leq N} = A_{\lambda+\delta_N}(x). \end{align*} Since $A_{\delta_N}(x)$ is a nonzero polynomial in the integral domain $\mathbb{Z}[x_1,\dots,x_N]$, cancellation in the fraction field gives \begin{align*} [t^{\lambda+\delta_N}]K(x,t) = \frac{A_{\lambda+\delta_N}(x)}{A_{\delta_N}(x)} = s_\lambda(x). \end{align*} Hence $D_{\lambda,N}(x)=s_\lambda(x)$. Since $D_{\lambda,N}(x)=J_{\lambda,N}(x)$, we obtain \begin{align*} J_{\lambda,N}(x)=s_\lambda(x). \end{align*}[/step]

[guided]We now connect the coefficient formula to the finite Schur polynomial without leaving an unproved antisymmetrization assertion. First extend $\lambda$ by zeros: for $r<b\leq N$, set $\lambda_b=0$. This lets us form the $N$ by $N$ polynomial-valued map \begin{align*} D_{\lambda,N}: \mathbb{Z}^N &\to \mathbb{Z} \\ x &\mapsto \det\bigl(h_{\lambda_i-i+j}(x)\bigr)_{1 \leq i,j \leq N}. \end{align*} Equivalently, $D_{\lambda,N}(x)$ denotes this determinant as an element of $\mathbb{Z}[x_1,\dots,x_N]$. Why does this not change the original $r$ by $r$ determinant? If $i>r$ and $j<i$, then $\lambda_i-i+j=j-i<0$, so the entry is $h_{j-i}(x)=0$. If $i=j>r$, the entry is $h_0(x)=1$. Thus the lower-right block is upper triangular with diagonal entries $1$, and the lower-left block is zero. Therefore \begin{align*} D_{\lambda,N}(x)=J_{\lambda,N}(x). \end{align*} Define \begin{align*} \delta_N:=(N-1,N-2,\dots,0). \end{align*} Applying the coefficient extraction identity already proved, now with $N$ rows and variables $t_1,\dots,t_N$, gives \begin{align*} D_{\lambda,N}(x) = [t^{\lambda+\delta_N}] \left( \prod_{i=1}^{N}H_x(t_i)A_{\delta_N}(t) \right). \end{align*} Here $[t^{\lambda+\delta_N}]$ means extraction of the coefficient of \begin{align*} t_1^{\lambda_1+N-1}t_2^{\lambda_2+N-2}\cdots t_N^{\lambda_N}. \end{align*} Using the complete homogeneous generating series, this becomes \begin{align*} D_{\lambda,N}(x) = [t^{\lambda+\delta_N}] \left( A_{\delta_N}(t)\prod_{i=1}^{N}\prod_{a=1}^{N}(1-x_a t_i)^{-1} \right). \end{align*} In finite variables, we use the standard characterization of the Schur polynomial by the [bialternant formula](/page/Schur%20Polynomial), equivalently to the tableau definition of $s_\lambda(x)$: \begin{align*} s_\lambda(x):=\frac{A_{\lambda+\delta_N}(x)}{A_{\delta_N}(x)}, \end{align*} where \begin{align*} A_{\lambda+\delta_N}(x)&:=\det\bigl(x_a^{\lambda_b+N-b}\bigr)_{1 \leq a,b \leq N},\\ A_{\delta_N}(x)&:=\det\bigl(x_a^{N-b}\bigr)_{1 \leq a,b \leq N}. \end{align*} We now prove the needed alternant coefficient computation instead of citing it. Define the formal power series-valued map \begin{align*} K: \mathbb{Z}^N \times \mathbb{Z}^N &\to \mathbb{Z}[[t_1,\dots,t_N]] \\ (x,t) &\mapsto A_{\delta_N}(t)\prod_{i=1}^{N}\prod_{a=1}^{N}(1-x_a t_i)^{-1}, \end{align*} where $t=(t_1,\dots,t_N)$. The goal is to show that the coefficient of $t^{\lambda+\delta_N}$ in $K(x,t)$ is $s_\lambda(x)$. The determinant identity behind this is \begin{align*} A_{\delta_N}(x)K(x,t) = \det\left((1-x_a t_i)^{-1}\right)_{1 \leq a,i \leq N}. \end{align*} To verify it, multiply both sides by the common denominator $\prod_{a=1}^{N}\prod_{i=1}^{N}(1-x_a t_i)$. The resulting polynomial on the determinant side is alternating in the $x$ variables because swapping two $x$ variables swaps two rows, and it is alternating in the $t$ variables because swapping two $t$ variables swaps two columns. For each fixed $a$, the degree in $x_a$ is at most $N-1$, since the row indexed by $a$ contains one factor from each column except the selected reciprocal denominator; likewise, for each fixed $i$, the degree in $t_i$ is at most $N-1$. Therefore divisibility by $A_{\delta_N}(x)A_{\delta_N}(t)$ forces the quotient to have degree $0$ in every variable, so the quotient is constant. Comparing the coefficient of $x_1^{N-1}x_2^{N-2}\cdots x_N^0t_1^{N-1}t_2^{N-2}\cdots t_N^0$ gives this constant as $1$, which proves the identity. Now expand the determinant in the formal power series ring. Since \begin{align*} (1-x_a t_i)^{-1}=\sum_{k=0}^{\infty}x_a^k t_i^k, \end{align*} multilinearity in the columns gives \begin{align*} \det\left((1-x_a t_i)^{-1}\right)_{1 \leq a,i \leq N} = \sum_{k_1,\dots,k_N \geq 0} \det\bigl(x_a^{k_i}\bigr)_{1 \leq a,i \leq N} \prod_{i=1}^{N}t_i^{k_i}. \end{align*} Extracting the coefficient of \begin{align*} t^{\lambda+\delta_N}=t_1^{\lambda_1+N-1}t_2^{\lambda_2+N-2}\cdots t_N^{\lambda_N} \end{align*} therefore gives \begin{align*} [t^{\lambda+\delta_N}]\bigl(A_{\delta_N}(x)K(x,t)\bigr) = \det\bigl(x_a^{\lambda_i+N-i}\bigr)_{1 \leq a,i \leq N} = A_{\lambda+\delta_N}(x). \end{align*} Because $A_{\delta_N}(x)$ is a nonzero polynomial in the integral domain $\mathbb{Z}[x_1,\dots,x_N]$, we may cancel it in the fraction field. Thus \begin{align*} [t^{\lambda+\delta_N}]K(x,t) = \frac{A_{\lambda+\delta_N}(x)}{A_{\delta_N}(x)} = s_\lambda(x). \end{align*} Therefore \begin{align*} D_{\lambda,N}(x)=s_\lambda(x). \end{align*} Since $D_{\lambda,N}(x)=J_{\lambda,N}(x)$, this proves \begin{align*} J_{\lambda,N}(x)=s_\lambda(x) \end{align*} in $\mathbb{Z}[x_1,\dots,x_N]$.[/guided]

[step:Pass from finite alphabets to the ring of symmetric functions] The equality \begin{align*} \det\bigl(h_{\lambda_i-i+j}(x_1,\dots,x_N)\bigr)_{1 \leq i,j \leq r} = s_\lambda(x_1,\dots,x_N) \end{align*} holds for every $N \geq r$. Both sides are stable under adding a new variable and then setting it equal to $0$: for every $N \geq r$, \begin{align*} h_m(x_1,\dots,x_N,0)=h_m(x_1,\dots,x_N) \end{align*} for all integers $m$, and \begin{align*} s_\lambda(x_1,\dots,x_N,0)=s_\lambda(x_1,\dots,x_N). \end{align*} Therefore the finite-variable identities define the same stable symmetric function. Passing to the inverse limit defining the ring of symmetric functions gives \begin{align*} s_\lambda=\det\bigl(h_{\lambda_i-i+j}\bigr)_{1 \leq i,j \leq r}. \end{align*} This is the Jacobi-Trudi identity. [/step]