# How to convert $(k+1)!\times(k+2)−1$ to $(k+2)!−1$?

First of all, is it called "convert" in English or there is a proper word for this purpose?

$(k+1)!\times(k+2)−1$ to $(k+2)!−1$

And what is the rule for this step?

Best wishes,

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...the definition of factorial states that $(n+1)!=n!(n+1)$ – Alex Becker Apr 3 '12 at 3:38
You remember the definition of factorial. – Arturo Magidin Apr 3 '12 at 3:38
"convert" is OK, if a bit unusual in this context. Instead of "convert $P$ to $Q$" I'd probably say "show $P$ is equal to $Q$". – Gerry Myerson Apr 3 '12 at 4:23
How to convert 1+1 in 2 ? – Student Apr 3 '12 at 4:42
@Student: That proof takes some 300 pages or so... – The Chaz 2.0 Apr 3 '12 at 5:02

$$\rm \color{Blue}{(k+1)!}\cdot \color{Green}{(k+2)}=\color{Blue}{1\cdot2\cdot3\cdots k\cdot(k+1)}\cdot \color{Green}{(k+2)}=(k+2)!$$

There's no special "rule" at play above; if you know the definition of the factorial this is immediate.

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Nice touch with the colors. I didnt know you could do that! – Daniel Montealegre Apr 3 '12 at 3:43
At least something useful from this post. – Salech Alhasov Apr 3 '12 at 4:09

Without need, $k$ was assumed to be an integer here. One can instead use this definition $$z!=\Gamma(z+1)=\int_0^\infty t^{z} e^{-t}\, \mathrm{d}t. \!$$ which converges absolutely, if the real part of the complex number z is positive $(\Re(z) > 0)$, to show that $$(k+2)!=\Gamma(k+3)=\int_0^\infty t^{k+2} e^{-t}\, \mathrm{d}t=\underbrace{\left[-t^{k+2}e^{-t}\right]_0^\infty}_{=0}+(k+2)\int_0^\infty t^{k+1} e^{-t}\, \mathrm{d}t =(k+2)\Gamma(k+2)=(k+2)\cdot (k+1)!$$

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For any positive integer $n$, $n!$ is defined as, $$n!=n(n-1)\cdots1.$$

From the above definition it follows that $n!=n(n-1)!,\forall n\in\mathbb{Z^+}.$

Hence it follows that $(k+1)!\times(k+2)−1=(k+2)!−1.$

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