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?
 A: 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.
A: 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 $0^0=1$.  
A: Dan, the proof of BT indeed assumes 0^0=1.  The problem is that:
(a) there are many places where we assume 0^0 = 1,  and
(b) people do not acknowledge (a).
Calculus textbooks usually say that 0^0 is undefined, but at the same time, they contain dozens of formulas that assume 0^0=1.
Textbooks should honestly admits that it is standard practice to evaluate 0^0 to 1.
The argument for "undefined" is based on the mistaken belief that 0^0=1 leads to contradictions. It does not (if it did, we would have found out long ago, because the assumption 0^0=1 is used in subtle ways in lots of places).
But the real reason that I feel strongly that 0^0 must be 1 is because the "undefined" idea is contrary to a two deeply held convictions, namely (1) that the most convenient definition must be the best one, and (2) math is consistent, so given the fact that set theory tells us 0^0=1, it means that it is OK to use that everywhere.
A: We can make sense of $0^0=1$ in this way also : Expand $f(x)=a^x$ as Taylor series for real values of x and a is fixed positive real number and let a tends to zero. Its not rigorous but still supporting our assumption.
A: 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.
