Presumably this is a well-known identity. Does it have a name?pic.twitter.com/tsMBGLZGAk
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Presumably this is a well-known identity. Does it have a name?pic.twitter.com/tsMBGLZGAk
The Stirling numbers are the change of basis coefficients for changing between the power basis and the falling factorial polynomials basis for the space of polynomials. This is essentially one side of this fact.
This is what @MathPrinceps was trying to tell me yesterday, I think. I’m still not entirely sure I understand the connection. 
I understand how Stirling numbers relate the power basis to the falling factorial basis, but not what that has to do with this identity. I assume it’s something very obvious that I’m overlooking.
Sorry to be obscure, Robin. Let's just start at the beginning, then, to be sure of remaining clear throughout. Suppose we write x(x - 1) ... (x - n) = x^(n+1) - S(n, 1)x^n + S(n, 2)x^(n-1) - ... +/- S(n, n)x^1. Your identity is then a statement about these S coefficients.
For of course each S(n, k) is a sum of all the k-fold products of the numbers from 0 to n -- that is, S(n, k) is the kth elementary symmetric polynomial in n+1 variables, evaluated at 0, 1, ... , n. And of course those terms whose factors include 0 contribute nothing to the sum.
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.
And that is how I am accustomed to think of the Stirling numbers of the two different kinds -- that is, I am accustomed to regard them as *defined* by their respective roles as matrix elements in these two (mutually inverse) basis-change matrices.
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