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I have referred to many websites online regarding the proof, but I haven't understood it at all.. Please do help.

As far as I know, $n! = 1 \times 2 \times 3 \times 4 \times \dots \times n$

Then using the same logic, $0!=0$

Then why is it that $0!=1$ ?

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marked as duplicate by Omnomnomnom, gt6989b, Amzoti, Julian Kuelshammer, Ataraxia Aug 27 '13 at 16:27

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I have two explanation

1= as we know $(n+1)! = n!(n+1)$ so from here $n! =\frac{(n+1)!}{(n+1)}$

so we have $4!=4*3*2*1=24$ and $3!=\frac{4!}{4}=6$ and $2!=\frac{3!}{3}=2$ and $1!=\frac{2!}{2}=1$and so $0!=\frac{1!}{1}=1$

2= one simple way of understanding what the factorial means is to say: "given a set of n objects, n! is the number of different ways to arrange those objects." This makes sense for, for example, n=3: there are six different ways to arrange a set of three objects (try it yourself and see!) But there is only one way to arrange a set of 0 objects, since there is nothing to rearrange. If we didn't set 0! = 1, this description wouldn't work.

Another simple formula is, for n > m, n!/m! = (n-m)!. This works if 0! = 1: n!/0! = n!/1 = n! = (n-0)!. But if we had any other value for 0!, such as 0, this formula would make no sense and we'd have to change it to say "n!/m! = (n-m)! unless m = 0", which is a lot longer and less beautiful!

There are countless other examples of why 0! = 1 is useful. Can you think of any?

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    $\begingroup$ $n!/m!=(n-m)!$ is almost always false. $\endgroup$ – walcher Aug 27 '13 at 15:55
  • $\begingroup$ it is mentioned there read it $\endgroup$ – Dimensionless Aug 28 '13 at 4:51
  • $\begingroup$ $${n! \over m!(n-m)!}= {n\choose m}=1 \Longleftrightarrow m=n, 0$$ $\endgroup$ – walcher Aug 28 '13 at 8:28
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Here's one way to look at it– we notice that the factorial follows this rule: $$ \begin{align} n! &= n\times(n-1)\times(n-2)\times(n-3)\times\cdots\times1\\ &=n\times\left[(n-1)\times(n-2)\times(n-3)\times\cdots\times1\right]\\ &=n\times(n-1)! \end{align} $$ or, in other words, $$ (n-1)! = \frac{n!}{n} $$ For example, we note that $$ 2!=\frac{3!}{3}=\frac{3\times2\times1}{3}=2\\ 1!=\frac{2!}{2}=1\\ $$ With $n=0$, the same rule should hold. So, we define $0!$ as $$ 0!=\frac{1!}{1}=1\\ $$ Note that this doesn't work so well if you keep going down to negative numbers.

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I indicated, in a comment on the question, why $0!$ should, "using the same logic," be the product of no factors. But how should this product be defined? Here's one explanation of that. In general, if you form a product of $k$ factors, form another product of $l$ factors, and then multiply those two products, the result is the same as if you multiply all $k+l$ of the factors, that is, $$ (a_1a_2\cdots a_k)\cdot(b_1b_2\cdots b_l)=(a_1a_2\cdots a_kb_1b_2\cdots b_l). $$ This seems to be so basic a property of multiplication that we'd want to preserve it if we extend the notion of "product" to allow the case of no factors. So let's look at this property with $k=0$. It says that $$ (\text{product of no factors})\cdot(b_1b_2\cdots b_l)=(b_1b_2\cdots b_l). $$ The only way this can work, for general $b$'s, is if the product of no factors is defined to be $1$. (We'd get the same conclusion by taking $l=0$ instead of $k=0$.)

P.S. The same argument applies not only to multiplication but to any associative operation with an identity element: The operation applied to no inputs should produce the identity element. Thus, for example, the sum of no terms should be defined to be $0$, and the union of no sets should be defined to be the empty set.

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Another way to look at it:

Take sums for example:

$$e = \sum_{k=0}^{\infty} \frac{1}{k!}$$

Now, if $0! = 0$ then, $k = 0 \implies \frac{1}{0} = \widetilde{\infty}$

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