# Factorial equation: why does $n!\cdot(n + 2)! = n\cdot(n + 2)!\cdot(n - 1)!$ [closed]

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Daniel W. Farlow, user1551, 6005, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

## 2 Answers

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

• So easy but I couldn't see it. Thanks a lot :) – DimChtz Aug 25 '16 at 19:50
• This assumes n > 0 right? – jlars62 Aug 25 '16 at 21:03
• @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. – Alex Ortiz Aug 25 '16 at 21:16

You have

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

so

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

Therefore

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

And that's it.