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I got to the point in my proof by induction of (k+1+1)! -1 but I don't know where to go from there.

Prasun Biswas
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Adam
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    Do you mean $\sum_{j=1}^n j\cdot j!=(n+1)!-1$? – Prasun Biswas Feb 03 '21 at 22:35
  • Yes thank you for correcting that I still have issues with my formatting haha – Adam Feb 03 '21 at 22:36
  • If yes, then note that $(n!-1)+n\cdot n!=n!(n+1)-1=(n+1)!-1$ – Prasun Biswas Feb 03 '21 at 22:37
  • An alternative to induction is to note that the right side counts the number of non-identity permutations of $1,2,\ldots,n+1$. The left side counts the same thing: there are $n\cdot n!$ permutations in which $n+1$ is the largest number moved by the permutation, $(n-1)\cdot(n-1)!$ permutations in which $n$ is the largest number moved by the permutation, and, in general $j\cdot j!$ permutations in which $j+1$ is the largest number moved by the permutation. – Will Orrick Feb 03 '21 at 23:13
  • Also note that your expressions both represent the largest $(n+1)$-digit number in the factorial number system. – Will Orrick Feb 03 '21 at 23:18

2 Answers2

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Hint: the induction step is just: $$ (n+1)\cdot(n+1)!+(n+1)!-1=[(n+1)+1]\cdot(n+1)!-1=(n+2)!-1. $$

yurnero
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you can use this $$j.j! = (j+1)! - j! $$

Hassan
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