Why is a raised to the power of Zero is 1? Why is $a^0=1$  $\forall a \in Z, a\neq0$. I understand $2^4=2\cdot2\cdot2\cdot2$ How can I express $a^0$. I am serious about the practical proof of this
 A: If you want $x^n\cdot x^m=x^{n+m}$ then $x^0\cdot x^n=x^n$. This means that $x^0=1$.
Of course you are free to define it otherwise. But this gives a pretty good motivation why $x^0=1$.
A: when a nonzero number is divided by itself, the result is $1$. Therefore, for example, 
$$2^4\div 2^4=1$$
on the other hand, by the rule of powers, one knows that, for example, $$2^7\div 2^5=2^{7-5}=2^2$$
comparing these two notes we obtain that
$$1=2^4 \div 2^4=2^0$$
hence
$$2^0=1$$
A: The answer is simpler than one might expect: it is so by convention.
To understand why the convention is such and not something different, look at this "puzzle": what is the most logical way to continue the sequence 16, 8, 4, 2, ... ? Clearly, the numbers are always halved, so the logical next one is 1. Now, since the sequence is $2^4, 2^3, 2^2, 2^1$, it is only logical to assign $2^0$ the value $1$.
It turns out that this choice is a good one, for example because powers obey the laws you expect them to obey, such as $a^{mn} = (a^m)^n$ and $a^{m+n}=a^m\cdot a^n$ for each $a \neq 0$ and each whole $n, m \geq 0$.
A: \begin{align}
  a^0 &= a^{n-n} = a^na^{-n} = a^n/a^n = 1
\end{align}
A: You can think of it as an empty product, because $1$ is the multiplicative identity.
A: $
2^4 = 16.
$
Divide by $2$ to get $2^3 = 8$.
Divide by $2$ to get $2^2=4$.
Divide by $2$ to get $2^1 = 2$.
Divide by $2$ to get $2^0 = \text{?}$
