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This question already has an answer here:

What is the definition of raising a number to the zeroth power ($x^0$)? I know that many people say that "anything raised to the zeroth power is one" but this is clearly not true since $0^0$ is $undefined$. How then do mathematicians define $x^0$ such that for all real numbers not equal to $0$, $x^0=1$?

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marked as duplicate by 6005, Community Oct 12 '15 at 3:47

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  • $\begingroup$ I think it may have to do with the fact that 1 is the multiplicative identity of a field. I'm commenting for a notification . $\endgroup$ – Yunus Syed Oct 12 '15 at 3:43
  • $\begingroup$ We mathematicians bend the rules and change them anyway we want:) Ok, essentially, we accept $x^0=1$ for all $x$ not zero There are different levels of proof for this. $\endgroup$ – imranfat Oct 12 '15 at 3:43
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    $\begingroup$ Actually, many mathematicians define $0^0=1$. Ter are strong reasons to do this - any empty product is $1$. The product of $0$ 1's is the same as the product of $0$ 0's. $\endgroup$ – Thomas Andrews Oct 12 '15 at 3:47
  • $\begingroup$ You may find this interesting:askamathematician.com/2010/12/… $\endgroup$ – NoChance Oct 12 '15 at 3:58
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Depends how strong of a proof you are looking for, but one way you could think of it is $$x^0 = x^{1-1} = \frac{x^1}{x^1} = \frac{x}{x} = 1$$ It may not intuitively be equal to $1$, but it is necessarily equal to $1$.

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$$ x^n x^0 = x^{n+0} = x^n$$ Hence $x^0 = 1$.

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  • $\begingroup$ That's why one defines it like this ... $\endgroup$ – martini Oct 12 '15 at 4:17

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