What values of $0^0$ would be consistent with the Laws of Exponents?

I am using the following fundamental properties of exponentiation on $N$ as as basis for this discussion:

(1) $0^1 = 0$

(2) $\forall x\in N (x\ne 0 \implies x^0 = 1)$

(3) $\forall x,y\in N (x^{y+1}=x^y\cdot x)$

Missing, of course, is a value for $0^0$. But only $0^0=0$ or $1$ are consistent with the Laws of Exponents:

(4) $\forall x,y,z\in N (x^{y+z}=x^y\cdot x^z)$

(5) $\forall x,y,z\in N (x^{y \space\cdot z}=(x^y)^z)$

EDIT:

From (5), we must have $(0^0)^2=0^{0\times 2}=0^0$. Therefore, $0^0= 0$ or $1$. Is this correct?

Is there any way to eliminate $0$ (or $1$) as a possible value, with reference to the fundamental properties or the laws of exponents?

Solution 1:

If $c,d$ are cardinal numbers, then $c^d$ is the cardinal of the set of maps $d \to c$. This works for all cardinal numbers and the usual arithmetic laws hold. For $d=0$ we get $c^0=1$ since there is a unique map $\emptyset \to c$. This holds for all $c$, in particular $0^0=1$. So there is actually no debate what $0^0$ is or not, it is $0^0=1$ by the general definitions. No case distinctions are necessary. Forget about $0^0=0$, this is nonsense.

Solution 2:

Ok, this may be glib, but following up on the comment of Will Jagy above, consider the following:

$$ 1 = 1^n = (1+0)^n = \sum_{i=0}^{n} \binom{n}{i} 1^i \cdot 0^{n-i}. $$

Do you see what I'm getting at? What sense do we want to make out of the last term, $1^n \cdot 0^0$?

Actually, I need to clarify why I posted this as an "answer." Since the OP is looking to define $0^0$ in a way that stays consistent with "fundamental properties of exponents," and I feel that the Binomial Theorem is fundamental enough to qualify as essential to arithmetic, the example above rules out the possibility of $0^0 = 0$ while reinforcing $0^0=1$.