Why is $n^0 = 1$? Why is any number to the zeroeth power equal to 1?  I would think it would be equal to zero, since nothing multiplied by nothing is, well, I would think 0.  But it is 1?
Examples:
$(-5)^0 = 1$;
$0^0 = 1$;
$5^0 = 1$;
 A: Since $n^k=n^{k+0}=n^k \cdot n^0$. This short computation suggests that $n^0$ should be $1$.
A: Expanding on what has already been said, I want to emphasize that $a^0=1$ is a convention. There is no computation to be made, because there is no obvious meaning to "multiply with itself zero times". 
The reason why the convention is reasonable is what has been mentioned by Thorben's answer and Gregory's comment: the relation $a^{m+n}=a^ma^n$ is so nice that it makes sense to extend to the rest of the integers. Once you have $a^0=1$, you also get $a^{-n}a^n=a^{n-n}=1$, so $a^{-n}=1/a^n$. 
A: $$\frac{n^x}{n^x}=1, n \neq 0$$ By laws of indices,
\begin{align*}n^{x-x}&=1\\
\implies n^0&=1\end{align*} 
So that's how we prove that $n^0=1$
A: The way I think about this (which may or may not be write) is that exponentiation (by integers) is viewed as 
$$ a^{n}=1\cdot a\cdot a\cdot ... \cdot a$$
where there are $n$ copies of $a$.  If $n=0$, then there is only the $1$.
You could also view $a^{n}$ as
$$a^{n} = \prod_{i=1}^{n}a.$$
Then, just note that the empty product is $1$... unlike the empty sum which is $0$.  Again, this only works when $n$ is an integer.
