@John #2, Another neat proof of that comes from the stirling transform. Basically, applying the "stretching map" \\( f \mapsto f(e^x - 1)\\) is equivalent to multiplying the taylor coefficients of \\( f \\) by a matrix consisting of stirling numbers of the second kind. What's fascinating to me about this is what seems to have nothing to do with combinatorics (the "stretching map") is actually governed by combinatorial data. I wrote more on this in the link below if you're curious: