TITLE: Why does $n^0 = 1$?
QUESTION [9 upvotes]: 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 $n^0 = 1$.

REPLY [4 votes]: 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.