Two Proofs that Σk (-1)k k!S(n,k)=(-1)n

By Doron Zeilberger

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Written: June 26, 2011.
This quickie was written after Dan Haran sent me the following question .
Added Jan. 18, 2012: Christophe Vignat commented that a yet simpler version of the first simple proof of the fact that
(-1)n = Σ (-1)k k! S(n,k) is a direct consequence of the fact that the Stirling numbers of the second kind are the connection coefficients between the monomials xn and the descending factorials (x)j = x(x-1)...(x-j+1). In other words
xn = Σ S(n,k) (x)k .
Taking x=-1 yields the identity.

I thank Christophe for his elegant remark, and remark in my turn that a trivial extension of my first proof (replace -1 by t) (x is already taken in the paper) yields the famous connection formula that he used.

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