A peculiar identity from the usual expansion of \((a-1)^r\) (Q1386785)

The paper considers the following summation: \(\sum^{a-1}_{i=0}{a\choose i} (a- i)^r(- 1)^i\) and evaluates it as \(0\) for \(1\leq r\leq a-1\), where \(a>1\) is a positive integer, \(a!\) where \(r= a\) any positive integer, and \(a!(1+ 2+\cdots+ a)\) if \(r= a+1\) where \(a\) is any positive integer. Note that the last term corresponding to \(i= a-1\) is mistaken in the abstract, \((a^a- 1)\) instead of the correct \({a\choose a-1}\). The summation considered in the paper is a standard textbook exercise to count the number of surjections from an \(r\)-element set to an \(a\)-element set by inclusion-exclusion. From here the results easily follow, since there are no surjections if \(r< a\), and if \(r\geq a\), then the number of surjections is known to be \(a!S(r, a)\), where \(S(r,a)\) is the Stirling number of the second kind. Note that \(S(a,a)= 1\) and \(S(a+ 1,a)= {a+1\choose 2}\).