The most basic definition of addition is: $a+b$ is the cardinality of a set that is a disjoint union of two sets of cardinalities $a$ and $b$. This matches the intended interpretation of addition for natural numbers and extends readily to $\mathbb Z$, $\mathbb Q$, $\mathbb R$, $\mathbb C$ by the usual constructions.
The most basic definition of multiplication is: $a\cdot b$ is the cardinality of the cartesian product of two sets of cardinalities $a$ and $b$. This matches the intended interpretation for natural numbers and extends readily to $\mathbb Z$, $\mathbb Q$, $\mathbb R$, $\mathbb C$ by the usual constructions.
The most basic definition of exponentiation is: $a^b$ is the cardinality of the set of maps from a set of cardinality $b$ to a set of cardinality $a$. Since there is exactly one map from $\emptyset$ to any other set (namley the inclusion map), we have $a^0=1$ for all cardinalities $a$, including $a=0$. On th other hand, for a non-empty set, there are no maps to the empty set, hence $0^b=0$ for all $b\ne 0$.
The basic rules of exponentiation are valid for this definition (as provable by simple set-theoretic arguments), e.g. $a^{b+c}=a^b\cdot a^c$, $(a\cdot b)^c=a^c\cdot b^c$, $a^{b\cdot c}=(a^b)^c$. This is all very nice and pretty and consistant and thus give s definition of exponentiation in $\mathbb N_0$. You have problems already when trying to extend the definition to $\mathbb Z$ (what is $0^{-1}$?). You have more problems when extending to $\mathbb Q$ (watch out for the error in $-2 = (-8)^{\frac13}=(-8)^{\frac 26}=((-8)^2)^{\frac16}={64}^{\frac16}=2$). Problems in the negative continue when trying to extend the definition to $\mathbb R$ by continuity (because $\frac{2n}{2n+1}\to 1$, we should have $(-1)^{\frac{2n}{2n+1}}\to -1$, in fact already the rational extension is not continuous at $(0,0)$), not to mention $(-1)^{\frac12}=?$). All this is the reason why the existence of a nice function $\exp\colon \mathbb C\to \mathbb C$ with the fundamental properties $\exp(x+y)=\exp(x)\cdot\exp(y)$ and $\exp(0)=1$ that resemble the above laws of exponentiation so well is taken as a way to define exponentiation in $\mathbb C$ with its help (but still with restrictions due to branching) - even the suggestive notation $e^z$ for $\exp(z)$ is so common!
However, this definition forces one to leave certain gaps in order to keep continuity. While one might choose some unusual branch cut other than the negative reals, $0^0$ will always be sacrificed.
In the end, what we want is persistance (that is: as few exceptions to theorems as possible). Therefore, in the $\mathbb C$ and probably already in $\mathbb Q$, it is preferable to leave $0^0$ undefined.
But when working e.g. with integers $0^0=1$ is preferable.
Note also the following example:
One writes polynomials e.g. as $P(X) = \sum_{k=0}^n a_k X^k$. Here it is conventional to use $X^0=1$, even if one computes $P(0)$. This is much more convenient than needing to write $P(X)=a_0 +\sum_{k=1}^n a_k X^k$.
Some might argue that $X\ne 0$ as element of the ring $\mathbb Z[X]$, therefore $X^0=1$ does not reference the problem whether $0^0$ is defined or not.
UPDATE: For all mathematical operations, it is relevant how they are defined.
If I take ZFC as "the" basis of most of mathematics, there is not a single axiom of it dealing with exponentiation at all. In fact, thee are not even axioms about addition or multiplication. Thus for example $2+2=4$ is not an axiom or postulate, it is simply a theorem (based on the definitions of "+" and of course of "2" and "4"). I have described above, how addition and multiplication are defined for elements of suitable sets of numbers. Contrary to these two operations, exponentiation as a few problems: There are several ways of defining it, essentially one is coming from exponentiation of ordinal numbers (and gives $0^0=1$) and one coming from the exponential and logarithm function and their defining power series (plus some conventions about how to treat multivaluedness of the logarithm in the complex case) which has difficulty at assigning a value to $0^0$. Thus while the first definition does not apply at all to general real or complex numbers and does assign e.g. the value $-1$ to the expression $e^{\pi i}$, the second definition does not assign a value to $0^0$. Thus under closer inspection there are diffeent operations that are known under the name of exponentiation (and the same name is justified and even motivated by the fact that both definitions agree where they both define a value). Thus it may depend on the context whether or not you assign a value to the expression $0^0$ (or maybe even which value you assign?). Either way $0^0=1$ is not an axiom, instead it is either part of a definition or a theorem follwoing from the specific definition of exponentiation.
Then again, you may consoider all definitions as extensions to a theory by introducing new symbols and axioms describing them. If this is how you look at it, then $0^0=1$ can be viewed as an independent axiom iff it is not part/consequence of the basic definition of exponentiation- see above.
