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I apologise for a repost, but my rep is not high enough to ask in a comment. But, in this question Simplify sum of factorials with mathematical induction I am confused how:

$$(n+1)!-1+(n+1)(n+1)! = -1+(n+1)!(1+(n+1))$$

I expect it's quite simple, but I couldn't for the life of me see the solution. Any help would be much appreciated.

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  • $\begingroup$ $(n+1)!-1+(n+1)(n+1)! = -1+(n+1)!(1+(n+1)) = (n+2)(n+1)!-1$ $\endgroup$ – Henry Jan 20 '15 at 12:28
  • $\begingroup$ Haven't you just rewritten the question? $\endgroup$ – Elis Jones Jan 20 '15 at 12:30
  • $\begingroup$ When I wrote my answer the question had $n$s and $k$s. But you are correct: the middle and right hand sides are essentially the left hand side rewritten and tidied up. $\endgroup$ – Henry Jan 20 '15 at 12:32
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First of all, the actual wording has only one free variable: $$(n+1)! -1+(n+1)\cdot(n+1)! = -1 + (n+1)!\cdot(1+n+1) = (n+2)! - 1$$ All that is used are the usual distributive law, note that $(n+1)!$ is an integer just like any other and the $!$ just binds the immediately preceeding integer, and the equation $$n! = n \cdot (n-1)!$$ Here is the same with some highlighting: $$\begin{align*} \color{red}{(n+1)!} \color{blue}{- 1} + (n+1) \cdot \color{red}{(n+1)!} & \stackrel{y = (n+1)!}= \color{red}y \color{blue}{-1}+(n+1)\cdot \color{red}y\\ & = \color{blue}{-1} + \color{red}y\cdot(1+(n+1))\\ & = \color{blue}{-1} + \color{red}{(n+1)!} \cdot (1 + (n+1)) \\ & = -1 +(n+1)! \cdot (n+2) \\ & = -1 + (n+2)! \end{align*}$$

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  • $\begingroup$ I see, thank you! I was over complicating it... again. $\endgroup$ – Elis Jones Jan 20 '15 at 12:29
  • $\begingroup$ I did like one of the answers which got deleted, it said let $y=(n+1)!$, then, $-1+(y+y(n+1))=-1+y(1+(n+1))$ $\endgroup$ – Elis Jones Jan 20 '15 at 12:52
  • $\begingroup$ @ElisJones Do you want me to incorporate this into my answer to make it more elaborate or is the coloring sufficient? $\endgroup$ – AlexR Jan 20 '15 at 12:55
  • $\begingroup$ I found that to be more a little more intuitive. But you're answer is nice as it is. $\endgroup$ – Elis Jones Jan 20 '15 at 13:04
  • $\begingroup$ @ElisJones I've added it to make it even nicer ;) $\endgroup$ – AlexR Jan 20 '15 at 13:08
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$n \cdot n! = (n+1-1)n!=(n+1)n!-n!=(n+1)!-n!$
Now, sum of all this taking $n=1,2,3,...$
$1 \cdot 1!+2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n!$$=(2!-1!)+(3!-2!)+(4!-3!)+...+((n+1)!-n!)=(n+1)!-1!$

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  • $\begingroup$ Your post is overlapping to the right and I don't see how it answers the question at hand. $\endgroup$ – AlexR Jan 20 '15 at 12:57
  • $\begingroup$ Sorry, I've just provided a new method to you. I didn't read whole the question, I understood you need a general method. $\endgroup$ – Hardey Pandya Jan 20 '15 at 12:59
  • $\begingroup$ That should actually be an answer to the linked question, not this one. $\endgroup$ – AlexR Jan 20 '15 at 13:00
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For example, to simplify $6!$, substitute $6$ for $n$ in the following equation:$$n!=n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot...$$$$6!=6\cdot(6-1)\cdot(6-2)\cdot(6-3)\cdot(6-4)\cdot(6-5)$$and now simplify:$$6!=6\cdot5\cdot4\cdot3\cdot2\cdot1$$but you don't have to including the $1$ because it's the same thing as not using it:$$6!=6\cdot5\cdot4\cdot3\cdot2$$Now evaluate it:$$30\cdot4\cdot3\cdot2$$$$120\cdot3\cdot2$$$$360\cdot2$$$$720$$Use the regular equation to help you with not just this, but any whole-number factorial.

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