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Just as the Fubini numbers are the coefficients of the exponential generating function $$\sum_{k=0}^\infty (e^x-1)^k = \frac{1}{2-e^x},$$
the exponential generating function for the alternating sum is $$\sum_{k=0}^\infty (1-e^{-x})^k = e^x.$$
Just as the Fubini numbers are the coefficients of the exponential generating function $$\sum_{k=0}^\infty (e^x-1)^k = \frac{1}{2-e^x},$$
the exponential generating for the alternating sum is $$\sum_{k=0}^\infty (1-e^{-x})^k = e^x.$$
Just as the Fubini numbers are the coefficients of the exponential generating function $$\sum_{k=0}^\infty (e^x-1)^k = \frac{1}{2-e^x},$$
the exponential generating function for the alternating sum is $$\sum_{k=0}^\infty (1-e^{-x})^k = e^x.$$
Just as the Fubini numbers are the coefficients of the exponential generating function $$\sum_{k=0}^\infty (e^x-1)^k = \frac{1}{2-e^x},$$
the exponential generating for the alternating sum is $$\sum_{k=0}^\infty (1-e^{-x})^k = e^x.$$
