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?
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.
Follow-up

It can be shown that there exists not just 1 or 2 "exponent-like" functions on N as I suggested here, but an infinite number of them. And from any one of them, we can derive the usual Laws of Exponents. For formal proofs, see Oh, the ambiguity! at my blog.