Why does $n^0 = 1$? Why is it that $n^0 = 1$?
I understand how $n^2 = n*n$ and how $n^1 = n$ but I can't understand why $n^0 = 1$.
 A: \begin{align*}
n^m &= n\times \ldots \times n\; (\textrm{for}\; m \in \mathbb{Z}_{>0})\\
&\vdots\\
n^2 &= n\times n\\
n^1 &= n\\
n^0 &= 1\\
n^{-1} &= \frac{1}{n}\\
n^{-2} &= \frac{1}{n\times n}\\
&\vdots\end{align*}
To go from $n^{m}$ to $n^{m+1}$, you multiply by $n$, and to go from $n^{m}$ to $n^{m-1}$, you divide by $n$.
A: When you do $m\cdot n$, you start from $0$ and keep adding $n$ as many times as $m$ says: so if $m=0$, you just have $0$ because you have nothing else to do; if $m=1$, you get $0+n$; if $m=2$ you do $0+n+n$. And so on.
Exponentiation is exactly the same, but with multiplication replacing addition and $1$ replacing $0$: for doing $n^m$ you start from $1$ and keep multiplying.
If $m=0$ you have $1$ and do nothing else; if $m=1$ you have $1\cdot n$; if $m=2$ you have $1\cdot n\cdot n$. And so on.
This is usually formalized in a recursive definition:
$$
0n=0,\qquad (m+1)n = mn+n
$$
for multiplication, that becomes
$$
n^0=1,\qquad n^{m+1}=n^m\cdot n.
$$
So, for instance,
$$
n^3=n^2\cdot n=(n^1\cdot n)\cdot n=((n^0\cdot n)\cdot n)\cdot n
=((1\cdot n)\cdot n)\cdot n
$$
and parentheses can be omitted.
Of course this has been recognized much time later than when exponentiation was introduced as a shorthand for “repeated multiplication“. The law of exponents
$$
n^{a+b}=n^a\cdot n^b
$$
is (almost) obvious when $a,b>1$; but it can extended to the exponents $0$ and $1$ so that it still holds by defining $n^1=n$ and $n^0=1$.
Note that nowhere it's assumed that $n>0$. It can be any “number”, as long as a neutral element form multiplication exists and multiplication is associative; indeed this can be done in any semigroup with identity (also called monoid).
The symmetry with multiplication could be seen better if we wrote $^mn$, but, alas, people started to write it the other way around.
A: HINT:
Using Exponent Laws(1) , (2)
$$n^{a+b}=n^a\cdot n^b$$
Setting $a=0,$  $$n^b=n^b\cdot n^0$$
Check when $n^b$ can be cancelled
A: Exponentiation satisfies the laws of exponents:  $a^{b+c} = a^ba^c$ . If we want this law to still be satisfied when we extend to the case $b=0$, we need to have  $a^{0+c} = a^0a^c$ , and therefore we need to have  $a^0 = 1$.
A: You can think of the rule that $\frac{x}{x}=1$ for any nonzero number $x$.
Then, using for example $n^2$, we get:
$$1=\frac{n^2}{n^2}=n^{2-2}=n^0$$
Hence it makes sense to have $n^0=1$.
