Tricky Exponents

Since $0!=1$ , $1000^{0!}=1000^1=\boxed{1000}$ .

In first time, we need to calculate 0!

1st demonstration:

In simple terms, the idea of the factorial is used to compute the number of permutations of arranging a set of n numbers.

For $n = 1$ :

The number of permutations is $1! = 1$ . $\left \{ 1 \right \}$

For $n = 2$ :

The number of permutations is $2! = 2$ . $\left \{ 1, 2 \right \} \left \{ 2, 1 \right \}$

For $n = 3$ :

The number of permutations is $3! = 6$ . $\left \{ 1, 2, 3 \right \} \left \{ 1, 3, 2 \right \} \left \{ 2, 1, 3 \right \} \left \{ 2, 3, 1 \right \} \left \{ 3, 1, 2 \right \} \left \{ 3, 2, 1 \right \}$

$\ldots$

Therefore, we can conclude that, for $n = 0$ :

The number of permutations is $0! = 1$ . $\left \{ \right \}$

2nd demonstration:

$n! = n * \left (n-1 \right ) * \ldots * 2 * 1 \\ n! = n * \left ( n - 1 \right )! \Rightarrow 1! = 1 * 0!$

But we know that $1! = 1$ , therefore $1! = 1 = 1 * 0! \Rightarrow 0! = 1$

Now, we know that $0! =1$ , so $1000 ^{0!} = 1000^{1} = 1000$ .

Therefore $1000^{0!} = 1000$ .

Combinatorics: you can read 0! as "in how many ways can you organize nothing?" One. Simple enough.

0! IS EQUAL TO 1 THEREFORE { 1000 }^{ 1 } = 1000

$n! = (n - 1)! \times n$

Subbing n = 1,

$1! = 0! \times 1$ ,

$\frac {1!}{1} = 0!$

$0! = 1$

Therefore, $1000^{0!} = 1000$

The zero indicates the number of times it should by multiplied by the original number. Since it was 0, then that means it wasnt multiplied, so it was still 1000

0!=1 (There's only one arrangement of zero items) 1000^1=1,000

as 0!=1 and so (x)^1=x

0!=1. so,1000^0!=1000

Since 0 factorial is equal to 1 . We simply get 1000^1 as 1000

As 0! = 1 and 1000 to the power 1 = 1000