Prove that $0!=1$

$0!=1$

$1!=1$

$2!=2\times 1$

$3!=3\times 2\times 1$

$4!=4\times 3\times 2\times 1$

$~~~~$ $\vdots$

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

$(n-1)! = \dfrac{n!}{n}$

Putting $n =1$

$(1-1)! = \dfrac{1!}{1}$

$0! = 1$ [Proved]

Munem Shahriar - 3 years, 6 months ago

Another method of Proving this can be:

We know that the number of ways of selecting $r$ objects from $n$ distinct objects $(n \ge r)$ are $\displaystyle {n \choose r}= \dfrac{n!}{r!(n-r)!}$ .

We also know that the number of ways of choosing $n$ objects from $n$ objects is of course $1$ .

$\therefore$ $\dfrac{n!}{n!(n-n)!}=\dfrac{1}{0!}=1$ $\implies 0!=1$

Vilakshan Gupta - 3 years, 6 months ago

I think of $0!$ as being 'defined' as $1$ , rather than it being able to be proved that it is $1$ .

I think of $0!=1$ as being true as the factorial function is 'multiplicative' in the sense that it is all about multiplication, and $1$ is defined as the identity element of multiplication.

Arthur Conmy - 3 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Prove that 0!=10!=10!=1

Comments

Prove that $0!=1$