for $n$ an integer, why is $n^0=1$ ?? This is so going to cost me....  
I was wondering why for any integer $n$: $n^0 =1$.
Perhaps It's because $n$ is a round number and if $m$ is a non negative integer, also round then:
$$n^m = 1 \cdot n^m=1 \cdot \underbrace{n \cdot n \cdots n}_\text{m times}$$
So maybe if we set $m=0$ then:
$$n^m = n^0 = 1 \cdot n^0= 1 \cdot \underbrace{n \cdot n \cdots n}_\text{0 times} = 1$$
Maybe that is because an integer multiplied by it self $0$ times is really doing nothing.
Wait so maybe $0$ multipled by $0$, $0$ times....? $$0^0 = 1 \cdot 0^0 = 1 \cdot \underbrace{0 \cdot 0 \cdots 0}_\text{0 times} = 1 \space\space??$$ 
$QED.$ 
If someone could prove to me (I'm trying) that $|x|^{|x|}$ is continuous in $\mathbb{R}$... Because we already know $lim_{x\to0}|x|^{|x|}=1$..
 A: We can see that $\frac{a^n}{a^m} = a^{n-m}$. So if $n = m$, then we get $\frac{a^n}{a^n} = 1 = a^{n-n} = a^0$.
Of course this intuition fails in explaining what $0^0$ is.
A: The idea behind saying "if you multiply nothing you get $1$" is called the empty product, which comes up often in discrete math and abstract algebra.  A similar argument, for example, motivates the definition that $0!$ should be $1$.
This is one justification among several that $0^0$ should be defined as $1$.
Alternatively, for the case of a positive $n$: if you believe that $n^{a+b} = n^{a}n^b$ and that $n^{-m} = 1/n^m$, then certainly we should have
$$
n^0 = n^{1-1} = n^1n^{-1} = n/n = 1
$$
so if $n$ is non-zero, we should certainly have $n^0 = 1$.
A: Your explanation of $n^0$ is perfectly fine.
For $0^0$ we can look at it in 2 diffrent ways:
1) any number to the power 0 is 1 so $0^0$ should be 1
2) 0 to any power is just zero multiplied by itself some times so it should be 0  
The fact that diffrent ways to evaluate this give diffrent results makes this undifined, so $0^0$ is not equal to anything
edit: you can see this anayticly:
1) the function $y=x^0$ approaches 1 if we let x approach 0 from the right.
2) the function $y=0^x$ approaches 0 if we let x approach 0 from the right.
A: a^n is a multiplied by itself n number of times. When n is negative, it's 1 / a multiplied by itself n number of times. So if a^1 is a by itself and a^-1 is 1 / a, when you multiply them together to "add" the exponents to equal zero, you get a / a which = 1, even when a is 0
A: It is not obvious to look at but it follows from the following;
If we multiply a number raised to a power by the same number raised to another power we can add the powers to find the answer, for example
$2^2\cdot2^3=2^{(2+3)}=2^5$ 
$4 \times 8 = 32$
If we divide a number raised to a power by the same number raised to another power we can subtract the powers to find the answer, for example
$\frac{3^6}{3^4}=3^{(6-4)}=3^2$
$\frac{729}{81}=9$
We know that if we divide a number by itself we get 1 so
$\frac{a^3}{a^3}=a^{(3-3)}=a^0=1$
I hope this helps.
A: We can use the limit definition as well. As long as a limit is continuous on both sides then it is true.
What happens if we approach it from the positive side? Well, let's take a constant, say 10, to help us see. $10^1$ = 10. That's easy. $10^{1/10}$ is just $\sqrt[10]{10}$, or What number do you raise to the 10th power to get to 10? The answer is 1.2589... Let's try $10^{1/100}$, or $\sqrt[100]{10}$. What number do you raise to the 100th power to get to 10? The answer is 1.023... Let's try $10^{1/10000}$, or $\sqrt[10000]{10}$. What number do you raise to the 10000th power to get to 10? The answer is 1.0000... etc. So we're sure about the positive side. But what about the negative side? Let's figure it out.
$10^{-1}= 1/10.$ Ok. Let's keep going. $10^{-1/10} = \frac{1}{\sqrt[10]{10}}$, which just so happens to equal to 0.79... Let's skip right to the good stuff. $10^{-1/1000} = \frac{1}{\sqrt[1000]{10}}$. This is about .9977... so let's try $10^{-1/1000000} = \frac{1}{\sqrt[1000000]{10}}$ which is about 0.999997. You get the point. Therefore, the limit as x approaches 0 for all $x^0$ where x is not equal to 0 is 1.
