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how to prove that $$\sum_{k=1}^{m}k!k=(m+1)!-1$$ without induction ?

my only try is to put $k!=\Gamma(k+1)$ then use geometric series with some steps but I got complicated integral

If any one can solve it using my way or similar way using calculus technique

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  • $\begingroup$ First of all, all you can ever do with this sort of problem is hide the induction. You can't even define the sum "without induction." $\endgroup$ – Thomas Andrews Mar 12 '14 at 23:12
  • $\begingroup$ I personally want a counting argument. $\endgroup$ – abnry Mar 12 '14 at 23:16
  • $\begingroup$ @nayrb See my answer, then. $\endgroup$ – Thomas Andrews Mar 12 '14 at 23:21
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$$ \sum_{k=1}^{m}k!k \\ \sum_{k=1}^{m}k!(k+1-1) \\ \sum_{k=1}^{m}k!(k+1)-k! \\ \sum_{k=1}^{m}(k+1)!-k! \\ = 2!-1!+3!-2!+4!-3!+\cdots + (m+1)!-m! \\ =(m+1)!-1 $$

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  • $\begingroup$ oh my god but I need 4 minutes more $\endgroup$ – user130806 Mar 12 '14 at 23:17
  • $\begingroup$ Telescopy is induction in disguise. But as Thomas Andrews pointed, it is impossible to prove this without induction. This argument is kind of nice, though. +1. $\endgroup$ – chubakueno Mar 12 '14 at 23:31
  • $\begingroup$ @chubakueno I agree and thanks. $\endgroup$ – Priyatham Mar 12 '14 at 23:32
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There is a combinatorial approach.

The number $(m+1)!-1$ counts the number of non-identity permutations of $\{1,2,\dots,m+1\}$.

On the other hand $k\cdot k!$ counts the number of such permutations that fix all elements greater than $k+1$ but not $k+1$.

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