Let n be any integer

so,\(n!\)=\(1\times2\times3\times\dots\times (n-1)\times n\)

and, \(n-1!\)=\(1\times2\times3\times\dots\times (n-1)\)

now, multiplying and dividing R.H.S. of \(n-1!\) by n, we get

\[\frac { 1\times 2\times 3\times ....\times n-1\times n }{ n }\]

which is actually \[\frac { n! }{ n }\]

now, put \(n=1\) in \(n-1!\), we get

\(1-1!\)=\(0!\) on L.H.S. and \[\frac { 1! }{ 1 }\]=\(1\) in R.H.S.

so, \(0!\)=\(\boxed1\)

Please give some more solutions to this..Thank you!

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## Comments

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TopNewestIf n! is defined as the product of all positive integers from 1 to n, then: 1! = 1*1 = 1

2! = 1*2 = 2

3! = 1

23 = 64! = 1

23*4 = 24 ...n! = 1

23...(n-2)(n-1)nand so on.

Logically, n! can also be expressed n*(n-1)! .

Therefore, at n=1, using n! = \(n*(n-1)!\)

1! = 1*0! which simplifies to 1 = 0!

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The most comprehensible explanation by far...! Thank you :)

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You are welcome...

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Using words alone you can determine 0!=1. factorials are used to tell us how many ways there are to arrange n amount of objects. if there are 0 objects there is only 1 way to arrange it. The proof is a nice way to show it but for basic understanding of what a factorial is and does words can help people who are learning it for the first time.

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niceee....;)

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Very good explanation of 0!=1 by Yoogottam Khandelwal.

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Thank you!

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One can prove this using definition of permutations.

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Could you expand it?

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