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PRELIMINARY. The Cauchy identity for Schur polynomials reads $$ \sum_{\lambda}s_\lambda(x_1,...,x_n)s_\lambda(y_1,...,y_n) =\prod_{i,j=1}^n\frac 1{1-x_iy_j}, $$ where $s_\lambda$ are the Schur polynomials and the sum on the left-hand side runs over all partitions (of length $\leq n$).

Denoting $p_k(x_1,...,x_n)=x_1^k+...+ x_n ^k$, we also have the identity $$ \exp\left(\sum_{k\geq 1}\frac{p_k(x_1,...,x_n)y^k}{k}\right) = \prod_{i=1}^n \frac 1{1-x_iy}, $$ which implies that we can rewrite Cauchy identity as $$ \sum_{\lambda}s_\lambda(x_1,...,x_n)s_\lambda(y_1,...,y_n) =\exp\left(\sum_{k\geq 1}\frac{p_k(x_1,...,x_n)p_k(y_1,...,y_n)}{k}\right).\qquad (\star) $$

THE QUESTION. Let now $p_1,p_2,...$ be an infinite set of independent variables, and let $S_\lambda(p_1,p_2...)$ be the Schur functions, namely the expression of Schur polynomials in the power-sum polynomials. Can the identity $$ \sum_{\lambda}S_\lambda(p_1,p_2,...)s_\lambda(y_1,...,y_n) =\exp\left(\sum_{k\geq 1}\dfrac{p_k(y_1^k+...+y_n^k)}{k}\right)\qquad (\star\star) $$ be deduced directly from $(\star)$ by an $n\to\infty$ limit argument?

Maybe I am missing something about rigorously inferring identities in the ring of symmetric functions (which is the inverse limit as $n\to\infty$ of the rings of symmetric polynomials in $n$ variables) from identities in the rings of symmetric polynomials in $n$ variables.

Any help is greatly appreciated!

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