0's Exponents are impossible? I've had something that's been bugging me, and I tried research and asked my math teacher. None had sufficient answers. 
The concept of $0$ is that when $0$ goes to any exponent except for $0$, it becomes $0$. For example,
$0^3 = 0$, but
$0^0 =$ undefined
However, the proof that $0^0$ is undefined is shown thus:
$0^x$
... (divided by) = $0^{(x-x)} = 0^0$ = undefined
$0^x$
You can apply this to any exponent though, such as:
$0^6$
... = $0^6 = 0$ and $0^3 = 0$, so this expression is equal to $0/0$, which should be 
$0^3$ undefined, right?
Am I doing something wrong here? Please help!
Gil
 A: To clarify one thing (Michael already posted a great answer). You don't prove something is undefined. Undefined is not some mystical object. Undefined means we have not defined it. 
You can prove something can't be defined in a way consistent with other math. I can define $0^0$ anyway I like. I define $0^0=54$. There, now it is defined, but it is not consistent with other things. 
I won't add anymore as Michael already added enough.
A: One should not say "equals undefined"; one should say "is undefined".  The is the "is" of predication, not the "is" of equality.
$0^0$ is indeterminate in the sense that if $f(x)$ and $g(x)$ both approach $0$ as $x\to a$ then $f(x)^{g(x)}$ could approach any positive number or $0$ or $\infty$ depending on which functions $f$ and $g$ are.
But in some contexts it is important that $0^0$ be taken to be $1$.  One of those contexts is in the identity
$$
e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.
$$
This fails to hold when $z=0$ unless $0^0=1$, since the first term of the series is then $\dfrac{0^0}{0!}$.
In combinatorial enumeration it is also important in some contexts that $0^0$ is $1$, for the same reason $2^0$ is $1$: if you multiply by something $0$ times, it's the same as multiplying by $1$.  That is also one way of looking at the reason why $0!$ is $1$.
