In a previous post i was asking about a complicated sum which seems not possible to simplify: Is it possible to simplify this sum?
I want now to calculate the simpler sum: $$ S := \sum_{k=1}^{n} \binom{n}{k} (k-1)!$$
I have tried to use derivative method like that: I define $f(x):= \sum_{k=1}^{n} \binom{n}{k} x^{k-1}$ We have: $$ f(x)=\dfrac{1}{x} \sum_{k=1}^{n} \binom{n}{k} x^{k} = \dfrac{1}{x} ( (x+1)^n -1 )$$ And the successive derivatives give: $$f^{(1)} (x)= \sum_{k=1}^{n} \binom{n}{k} (k-1) x^{k-2}$$ $$f^{(2)} (x)= \sum_{k=1}^{n} \binom{n}{k} (k-1)(k-2) x^{k-3}$$... I would like to evaluate in 1 but i don't see a way to obtain the sum $S$ i want. Is it possible to simplify it (maybe with another method) ?
Thank you.