It is pretty easy actually. I’m not going by the definition here. But this is still a mathematically correct solution.

1!= 1

2!= 1×2

3!= 1x2x3 and so on..

But notice how 3!= 2!x3

So (n+1)!= n!x(n+1) correct?

Now substitute n as 0, we get

1!=0!x1

Therefore, 0!=1!=1

If you wanna know more about this, try googling gamma function for factorials. There’s an integral function which can be integrated for n=0 for x=0 to ∞

It is pretty easy actually. I’m not going by the definition here. But this is still a mathematically correct solution.

1!= 1

2!= 1×2

3!= 1x2x3 and so on..

But notice how 3!= 2!x3

So (n+1)!= n!x(n+1) correct?

Now substitute n as 0, we get

1!=0!x1

Therefore, 0!=1!=1

If you wanna know more about this, try googling gamma function for factorials. There’s an integral function which can be integrated for n=0 for x=0 to ∞

## Satyam 🔥🔥🔥

n = n x (n-1)

⟹ n! = n x (n-1)!

Substitute n as 1, we get

⟹ 1! = 1 x ( 1-1 )!

⟹ 1 = 0! { ∵ 1! = 1 }

## Therefore, 0! = 1

n = n x (n-1)

⟹ n! = n x (n-1)!

Substitute n as 1, we get

⟹ 1! = 1 x ( 1-1 )!

⟹ 1 = 0! { ∵ 1! = 1 }

Therefore, 0! = 1

## Shivansh Preet

## Mehul

DID YOU NOTICE this pattern

n! = (n+1)! ÷ (n+1) GOT IT ?

Following this pattern,

when we put n=0, we get :

0! = (0+1)! ÷ (0+1) = 1! ÷ 1 = 1 ÷ 1 = 1

→ 0! = 1 (proved)

2> Let us consider some examples :

4! = 4*3*2*1 = 4 * 3!

3! = 3*2*1 = 3 * 2!

2! = 2*1 = 2 * 1!

DID YOU NOTICE THE PATTERN?

n! =n * (n-1)! GOT IT ?

putting n = 1, we get

1! = 1 * (1–1)! = 1 * 0!

→ 0! = (1! / 1) = (1 / 1) = 1

→ 0! = 1 (proved)

## Prince kumar

DID YOU NOTICE this pattern

n! = (n+1)! ÷ (n+1) GOT IT ?

Following this pattern,

when we put n=0, we get :

0! = (0+1)! ÷ (0+1) = 1! ÷ 1 = 1 ÷ 1 = 1

→ 0! = 1 (proved)

2> Let us consider some examples :

4! = 4*3*2*1 = 4 * 3!

3! = 3*2*1 = 3 * 2!

2! = 2*1 = 2 * 1!

DID YOU NOTICE THE PATTERN?

n! =n * (n-1)! GOT IT ?

putting n = 1, we get

1! = 1 * (1–1)! = 1 * 0!

→ 0! = (1! / 1) = (1 / 1) = 1

→ 0! = 1 (proved)

## Suraj kumar

n x (n-1)

⟹ n! = n x (n-1)!

Substitute n as 1, we get

⟹ 1! = 1 x ( 1-1 )!

⟹ 1 = 0! { ∵ 1! = 1 }

Therefore, 0! = 1

## Sandeep Kumar yadav

It is pretty easy actually. I’m not going by the definition here. But this is still a mathematically correct solution.

1!= 1

2!= 1×2

3!= 1x2x3 and so on..

But notice how 3!= 2!x3

So (n+1)!= n!x(n+1) correct?

Now substitute n as 0, we get

1!=0!x1

Therefore, 0!=1!=1

If you wanna know more about this, try googling gamma function for factorials. There’s an integral function which can be integrated for n=0 for x=0 to ∞

## Ritik kumar

It is pretty easy actually. I’m not going by the definition here. But this is still a mathematically correct solution.

1!= 1

2!= 1×2

3!= 1x2x3 and so on..

But notice how 3!= 2!x3

So (n+1)!= n!x(n+1) correct?

Now substitute n as 0, we get

1!=0!x1

Therefore, 0!=1!=1

If you wanna know more about this, try googling gamma function for factorials. There’s an integral function which can be integrated for n=0 for x=0 to ∞