From what I gather, we can't just define $0^0$ to be $0$ or $1$ or $69$ or whatever, because $\lim\limits_{x\mathop\to0}0^x=0$ and $\lim\limits_{x\mathop\to0}x^0=1$. So $0^0$ is called indeterminate
But why can we define (say) $2^3$? How do I know that if $\lim\limits_{x\mathop\to c}f(x)=2$ and $\lim\limits_{x\mathop\to c}g(x)=3$ then $\lim\limits_{x\mathop\to c}f(x)^{g(x)}=8$ for all functions $f$ and $g$? Is this true and if so is there a proof?
You can define $x^n$ any way you want so it is certainly true that $2^3$ and $0^0$ can both be defined.
Some people claim that defining $0^0=1$ once and for all could lead to contradictions, but this claim is wrong. It is based on a general distrust of $0$, combined with faulty reasoning, such as: (a) arguments that require the impossible (requiring a discontinuous function to be continuous) or (b) arguments that use erroneous claims such as $0^x = 0$ for all $x$ (which is clearly wrong, try $x=-1$).
In math, every definition is based on convenience. The reason that the notation $x^4$ was introduced is because it is shorter than $xxxx$.
If $x^0$ wasn't defined as $1$ (for every $x$, including $x=0$) then that would be highly inconvenient. Because wherever you now see $x^n y^m$, you would have to replace that with: "if $n=0$ then $y^m$, and if $m=0$ then $x^n$, and if both are $0$ then $1$, and otherwise $x^n y^m$".
Nobody uses that kind of cumbersome descriptions, we simply write $x^n y^m$. The way the notation $x^n y^m$ is commonly used is only correct if we define $0^0$ as $1$. Fortunately, this definition is supported by many good arguments (e.g. the empty product rule, combinatorial arguments, set theory, etc., all of these produce the same conclusion) while the arguments against this definition use steps that would not be accepted in other contexts.
People use very strange logic to argue that $0^0$ should be undefined. For instance, the argument $0^x = 0$ for $x>0$. The logic behind that argument is that if a statement $P(x)$ is true for $x>0$ then it should also be true for $x=0$. This type of logic would be rejected in any area of mathematics except in the $0^0$ debate.
– Mark Feb 11 '17 at 15:34About 3/4 of the arguments against defining $0^0$ are based on the "continuity rule", which says that if $f$ is a function that is discontinuous at a point $p$, then you may not define $f(p)$.
Of course, no serious mathematician who studied after the 19'th century accepts the "continuity rule" (except in the context of $0^0$) so violating this "rule" is perfectly OK.
– Mark Feb 11 '17 at 19:24You certainly can define $0^0$. There is absolutely no limit on what you can define; the only thing that may happen is that your definition turns out not to be useful.
So you've made two examples of possible definitions of $0^0$. Let's first at the definition $0^0=69$. This definition would break the power laws (for example, $69 = 0^{0\cdot 2} \ne (0^0)^2 = 69^2$), without giving any benefit at all (I couldn't tell a single problem that wouldbe simplified by that definition).
On the other hand, the definition $0^0=1$ does simplify quite a few things, for example it ensures that the binomial formula $$(a+b)^n = \sum_{k=0}^n {n\choose k} a^k b^{n-k}$$ also works unchanged if $a$ or $b$ is $0$.
What however is not possible is to define $0^0$ in a way that $x^y$ is continuous for $x=0$ and $y=0$.
$2^3$ is simple. We know how powers work and can multiply 2 by itself 3 times. We could do that for most numbers instead of 2 and 3, just by multiplying one by itself the other number of times. You obviously know that. $0^0$ is a little bit trickier. You may ask why, and the answer is that $0^0$ is what we like to call a degenerate case. A degenerate case is when a pattern that works over a lot of examples falls apart, or degenerates. Sort of like how we can have $1/3$ and $1/2$ and $1/1$ but $1/0$ cant exist. $0^x$ becomes $0$ for almost all numbers. $x^0$ becomes $1$ for almost all cases of x. Both of these work for all non-zero x's but at $0$ it collapses and it isnt that simple. For a better explanation and several viewpoints on why $0^0$ should actually be defined as 1 see Why is $0^0=1$? a previosuly asked question that discusses it quite well.
As to the second part regarding $2^3$. Limits commute with functions if and only if the function is continuous at that point. So your question is equivalent to asking if $f(x,y)=x^y$ is continuous at the point $p=(2,3)$. The answer to that is: yes, that is true (to see this, use the equation $x^y=e^{ln(x) y}$, which holds in a suitable open neighborhood of $p$).
– Mark Feb 12 '17 at 22:56