Presumably this is a well-known identity. Does it have a name?pic.twitter.com/tsMBGLZGAk
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Did I really get that wrong? Sorry. OK, so you proceed inductively: first, write x^(n-1) as a sum of falling factorials (with Stirling number coefficients), and then get x^n from that sum by multiplying each of its constituent summands by the appropriate [(x - n) + n] factor.
In the case of x^4, it goes like this: x^4 = x[x^3] = x[x(x - 1)(x - 2) + 3x(x - 1) + x] = [x(x - 1)(x - 2)][(x - 3) + 3] + 3[x(x - 1)][(x - 2) + 2] + [x][(x - 1) + 1] = x(x - 1)(x - 2)(x - 3) + 6x(x - 1)(x - 2) + 7x(x - 1) + x. Does that clarify the idea?
At least this approach shows why the coefficients of the falling factorial polynomials appearing in the sum for x^n must satisfy the same recursion relation obeyed by the Stirling numbers. And since both the recurrence and the initial conditions match, the numbers must, as well.
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