Question

Prove that 0! = 1

Answer 1

Welcome to open study

Answer 2

Hi,
0! = 1 is true simply by its "DEFINITION. "

Answer 3

and if anybody proved it then he can win a nobel prize

Answer 4

1 = 1! = 1 * 0!, hence dividing both sides by 1 yields 0! = 1.

Answer 5

its the base of mathematics

Answer 6

ok it is the result of Gamma function
\[\Gamma[n+1)=n!\]
when n=0
then
\[\Gamma[1]=1\]

Answer 7

using the formula n! = (n-1)(n-2)(n-3)...*n

Answer 8

n! = (n-1)(n-2)(n-3)...*1
not n

Answer 9

typo

Answer 10

if it was n then 0! = 0

Answer 11

I think we can use here Permutation...

Answer 12

We all know:

\[\huge ^nP_r = \frac{n!}{(n-r)!}\]

Answer 13

So to find 0!,

Put r = n here:

\[\large ^nP_n = \frac{n!}{(n-n)!}\]

As: \(^nP_n = n!\),

So,

\[\large n! = \frac{n!}{0!}\]

Divide by n! both the sides,

\[\large 1 = \frac{1}{0!}\]

\[\implies \large \color{green}{0! = 1}\]

Answer 14

n! = n(n-1)!
1! = 1(1-1)!
1 = 1(0!)
0! = 1

Answer 15

Many ways to prove...

Answer 16

Agree just a try for asking the first time

Answer 17

Okay..

Answer 18

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