Question

For anyone who know the hyperfactorial, is it a way to prove that ula1" title="0^{0}=1" alt="0^{0}=1" align="absmiddle" class="latex-formula">? We can show that
$H(1)=1^{1}H(0)$

$H(0)=1$

Answer 1

I don't think that there's a way to "prove" that $0^0$ equals anything, since that quantity is undefined.

Moreover, the hyperfactorials are defined as

$H(n) = 1^1\cdot2^2\cdot3^3\cdot\ldots\cdot n^n$

So, claiming that $H(1)=1^1\cdot H(0)$ wouldn't be true, because $H(1)$ is already the case that solves the recursion.

Answer 2

Answer:

No, the Hyperfactorial is not a way to prove that.

0 to the 0 power is just undefined, but it cannot be 1.

Step-by-step explanation:

H(1)=1^{1}H(0)

The Hyperfactorial is defined as the result of multiplying a given number of consecutive integers from 1 to the given number, each raised to its own power.

No, since 1^1=1, H(1)—Consecutive Integer is 0.

0 is what you get in the end, no matter what.