How do I show the following property of a factorial?

How do I show the following?

$$\frac{n!}{(k+1)!(n-(k+1))!}=\frac{n-k}{k+1}\frac{n!}{k!(n-k)!} \text{ for } k=0,1,\ldots,n-1$$

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Use the definition of factorial. –  Chris Eagle Oct 10 '12 at 8:02
It s obvious, since $\frac{n-k}{(n-k)!}=\frac{1}{(n-(k+1))!}$ –  Babak S. Oct 10 '12 at 8:04
@BabakSorouh, why? –  idealistikz Oct 10 '12 at 8:11
@idealistikz: The Brian's is the complete answer. –  Babak S. Oct 10 '12 at 8:19
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2 Answers

It’s just algebra:

\begin{align*} \frac{n-k}{k+1}\cdot\frac{n!}{k!(n-k)!}&=\frac{(n-k)n!}{(k+1)k!(n-k)!}\\ &=\frac{(n-k)n!}{(k+1)!(n-k)!}\quad\text{since}(k+1)!=(k+1)k!\\ &=\frac{\color{red}{(n-k)}n!}{(k+1)!\color{red}{(n-k)}(n-k-1)!}\quad\text{since}(n-k)!=(n-k)(n-k-1)!\\ &=\frac{n!}{(k+1)!(n-k-1)!}\\ &=\frac{n!}{(k+1)!\big(n-(k+1)\big)!} \end{align*}

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@idealistikz: No; where do you think that there should be one? –  Brian M. Scott Oct 10 '12 at 8:23

$$\frac{n!}{(k+1)!(n-(k+1))!}=\frac{n!}{(k+1)k!(n-k-1)!}=\frac{n!}{(k+1)k!\frac{(n-k)!}{n-k}}=\frac{n-k}{k+1}\frac{n!}{k!(n-k)!}$$

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