Why isn't $\frac{0^{0!}}{0!^0}$ not undefined? I got into a big argument with my teacher about this. I am saying that it is undefined because every time I work it out, I end up getting $\frac{0}{0}$ which I know to be undefined. 
 A: Conventionally, $0!=1$ and $x^0=1~\forall x\neq0$. From this we obtain
\begin{align}
\frac{0^{0!}}{0!^0} = \frac{0^1}{1^0} = \frac{0}{1} = 0\,.
\end{align}
Hope this helps.
A: 
Hi there, you're wrong - you should apologise to your teacher

I don't know why this question has gotten so much attention, the OP needs to learn to read and not panic! There is no problem.
$0!=1$ the denominator of $0!^0$ is $1^0$ which is $1$
The numerator is $0^{0!}=0^1=0$
The only problem you could have is if you had $x^0$ with $x=0$, which I consider undefined.
Next.
Why is $0!= 0$?
There is something I was taught as "the axiom of choice" but it isn't, it has a different name, it states that:

Given a decision with m outcomes, and another with n outcomes regardless of the first decision then the number of ways to decide is exactly:
  $$mn$$

If you have $n$ things and you have to choose an ordering, you have $n$ choices for the first thing, $n-1$ for the second, $n-3$ for the third.... so forth. This is where the $n!$ definition comes from
How many ways are there to arrange a collection of 0 objects? One way.
There is only one way I can present you with nothing.
