Can someone explain how this:

$$n!\cdot(n + 2)!$$

could become:

$ n\cdot(n + 2)!\cdot(n - 1)!$


closed as off-topic by T. Bongers, Daniel W. Farlow, user1551, 6005, user223391 Aug 28 '16 at 13:54

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$$ n! = n \cdot (n - 1)! $$ That's all.

  • $\begingroup$ So easy but I couldn't see it. Thanks a lot :) $\endgroup$ – DimChtz Aug 25 '16 at 19:50
  • $\begingroup$ This assumes n > 0 right? $\endgroup$ – jlars62 Aug 25 '16 at 21:03
  • 1
    $\begingroup$ @jlars62, The gamma function defined by $\Gamma(t) = \int_0^\infty x^{t-1}e^{-x}\,dx$ satisfies the functional equation $\Gamma(t+1) = t\Gamma(t)$, and is itself an extension of the factorial function for nonnegative integers. See here: en.wikipedia.org/wiki/Gamma_function. There are definitions of the gamma function that make use of negative integers as well. However, my answer speaks to $n \gt 0$, yes. $\endgroup$ – Alex Ortiz Aug 25 '16 at 21:16

You have

$$n!=1\times 2\times\cdots\times n$$


$$n!=(1\times\cdots \times n-1)\times n=(n-1)!\times n.$$


$$n!\cdot (n+2)!=n\cdot (n-1)!\cdot (n+2)! = n\cdot (n+2)!\cdot (n-1)!.$$

And that's it.


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