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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 $n^0 = 1$.

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  • $\begingroup$ there is $$ n^k = \Pi^k_{i=1}i$$ but that relies on $$ \Pi^0_{i=1}i=0$$ $\endgroup$ – ratchet freak Apr 17 '14 at 11:54
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    $\begingroup$ Your formula describes $n!$ $\endgroup$ – Red Alert Apr 17 '14 at 18:39
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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 $a^0 = 1$.

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  • $\begingroup$ Thanks for explaining it simply :). $\endgroup$ – C1D Apr 17 '14 at 9:21
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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

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\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 $n$.

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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.

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    $\begingroup$ The only answer that actually says what's going on. $\endgroup$ – Git Gud Apr 17 '14 at 10:39
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    $\begingroup$ @GitGud You know, I'm really bored to see something like $n^0=n^{2-2}=n^2/n^2$ which is meaningless because it uses an extension of the law of exponents which can only be done after we have proved the basic law for nonnegative exponents. And this proof relies of course on defining $n^0=1$. $\endgroup$ – egreg Apr 17 '14 at 10:43
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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 $n^0=1$.

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    $\begingroup$ $x/x=1$ for any non zero number $x$? $\endgroup$ – RDizzl3 Apr 17 '14 at 8:49

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