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I understand that $$n! = n (n-1) (n-2)\cdots 2 \cdot 1.$$ My book says this can also be written as $$n (n-1)!$$ Without telling me why

My question is How and why is that? Why can't we leave it as it is? Basically looking for an easy to understand explanation.

Thank You

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2 Answers 2

up vote 12 down vote accepted

If you substitute $n-1$ into the definition, you get

$$(n-1)! = (n-1) \cdot (n-2) \cdot \dotso\cdot2 \cdot 1\;.$$

This is $n!$ except that the initial factor of $n$ is missing, so $n!=n(n-1)!$. In words: Since $n!$ is the product of all numbers up to $n$, it's the product of $n$ with the product of all numbers up to $n-1$.

As to why we can't leave it as it is, we can; I don't see how your book was implying that we can't. The reason it sometimes makes sense to rewrite the factorial like this is mainly that that's useful in inductive proofs, where some property involving a factorial $n!$ is reduced to the corresponding property involving the factorial $(n-1)!$.

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I substituted (n-1) and i found to my surprise it brings back the definition of n!.I'am convinced now.Thank you. –  alok Sep 12 '11 at 11:58

I want to add that understanding what factorials represent, combinatorially, is another way of understanding where the second formula 'comes from'. Remember that $n!$ is the number of ways of arranging $n$ distinct objects (say, pool balls numbered $1\ldots n$) in a line; now consider the ball labeled $n$. There are $n$ 'places' that it can go in the line; the first slot, the second slot, ..., all the way to the $n$th slot. And wherever you take it out from the line, the other $(n-1)$ balls can be arbitrarily arranged in $(n-1)!$ ways, so the total number of arrangements $n!$ must be the same as the product (places to put the $n$th ball)$\times$(ways of arranging the other balls) - or in other words, $n\times(n-1)!$.

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