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I have an equation containing factorials here

$$(k+1)!+(k+1)(k+1)!$$

yet I am having a hard time understanding how to simplify it using algebra. A simple search on wolfram gets me a reduced form of

$$(k + 2)! $$

This would be a great refresher to such problems, sadly I don't know the elementary operations to reduce it.

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    $\begingroup$ Hint: $a+(k+1)a=(k+2)a$. $\endgroup$ – dxiv Oct 13 '16 at 3:56
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\begin{align} &(k + 1)! + (k + 1)(k + 1)!\\ = & (k + 1)!(1 + k + 1)\\ = & (k + 1)!(k + 2)\\ = & (k + 2)(k + 1)(k)(k - 1)....(1)\\ = & (k +2)!\\ \end{align} First you factor out the $(k + 1)!$ and simplify $(1 + k + 1) = (k + 2)$. You are left with $(k + 2)(k + 1)!$ which is just $(k + 2)!$.

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  • $\begingroup$ I did not see that. Especially the last three steps, where you an simply remove K + 1 in the third step. $\endgroup$ – MooCow Oct 13 '16 at 4:41
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Factor out the $(k+1)!$:

$$(k+1)! + (k+1)(k+1)! = (1+k+1)(k+1)! = (k+2)(k+1)! = (k+2)!$$

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  • $\begingroup$ You are factoring out $(k+2)!$, isn't it? $\endgroup$ – Sum Mar 25 '17 at 3:36

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