Presumably this is a well-known identity. Does it have a name?pic.twitter.com/tsMBGLZGAk
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So the sum that appears on the left-hand side of your identity is indeed S(n, k). Which means that the only thing left is to show how the S(n, k) relate to the Stirling numbers. And it's clear from their definition that they relate the falling-factorial basis to the power basis.
So these S(n, k) are just the entries of the matrix inverse to that whose entries are the coefficients s(n, k) appearing in the expression for x^n in terms of the falling-factorial polynomials.
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