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$\bullet$ The $0!$ is something is pre-defined, and when even a calculator returns that it is because it is programmed to do so and doesn't actually do the calculation. It also cannot be expressed in the standard form of writing down the factories; i.e;

$$n!=n(n-1)(n-2)...2*1$$

Because if you do so you get $$0!=0$$ and also this is create an infinite series of products (But doesn't matter because $0*n=0$)

$\bullet$ And also if we look at this pattern that is; $$n! = \frac{(n+1)!}{(n+1)}$$

And aply this to $0!$ we get,

$$0!=1$$

So we could have chosen $0!=0$, but chose the latter. But I'm also not here to state that we could have chosen $0!=0$, but another similar problem.

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$\bullet$ Let's assume that $n^0=1$ for $n∈\Bbb R$. This is true for all numbers but when it comes to $n=0$, we get $$0^0 = 1$$

$\bullet$ Let's assume that $0^n=0$ for $n∈\Bbb R$. This is true for all numbers but when it also comes to $n=0$, we get $$0^0 = 0$$

So using two common laws in exponents to zero, gives contradictory results. This is much similar to the case that we have seen in $0!$ above. So we need to choose one to be pre-defined. But which one? Here we also have another entity;

$$\lim_{n\to 0}n^n=1$$.

So if $n=0.000001$, we get; $$0.000001^{0.000001} = 0.9999 ≈ 1$$

And also most calculators show up that $0^0$ is 1. But I have yet to see this used in conventional formulas and other places. The fact that I'm so adamant about this is that this can open up a lot of opportunities. For example, this could help in formulating a neat expression for the Signum function for which the existing one is;

$$sgn(n) = \begin{cases} \frac{|n|}{n}, & \text{if $n$ ≠ 0} \\[2ex] 0, & \text{if $n=0$} \end{cases}$$

So what are your opinions on this?

Rayreware
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    We can (and we do) define $0^0=1$. Simpler rationale: there is one mapping from the empty set to the empty set. – Angina Seng Sep 07 '19 at 04:55
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    I object to an empty product being $0$, hence your argument for $0!=0$ feels very wrong to me. – Vsotvep Sep 07 '19 at 04:56
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    If we have a sequence of numbers, and we take their product $c$, and we compare this with a new sequence that has adds more element $a$ to the original, then the new product should be $c\cdot a$. Hence if our original sequence was empty, we should have $c=1$ to preserve this property. – Vsotvep Sep 07 '19 at 04:57
  • @Vsotvep, yes I do agree on that and also $0!=1$, I was more curious on why this can't be conventionally done using $0^0$ too. EDIT; and also if we take can't apply the conventional $n!=n(n-1)(n-2)...$ with $0$ just like with negative integers, then why isn't $0! = undefined$ similar to that of the factorial of negative numbers? – Rayreware Sep 07 '19 at 05:07
  • You gave the answer yourself: "So using two common laws in exponents to zero, gives contradictory results." The value of $0!$ as defined by an empty product or by its semantical meaning (there is exactly one way to arrange an empty number of objects) is not contradictory, while $0^0$ has different values depending on point of view. – Vsotvep Sep 07 '19 at 05:13
  • @Rayreware The factorial of negative numbers makes no sense, since there are no such thing as negative sequences to take products over, or negative sets of elements to arrange – Vsotvep Sep 07 '19 at 05:15
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    There is snowball's chance in hell that a topic like $0^0$ would not have been discussed to death already. With a view of A) keeping all the discussion in the same place, B) avoiding rehashing of the same arguments, it is surely better to close this thread as a duplicate. My choice was simply the oldest and the most upvoted. Anyone having something new to say should, IMO, post there rather than here. – Jyrki Lahtonen Sep 07 '19 at 05:17
  • @Vsotvep, then if you go with $n!=n(n-1)(n-2)...2*1$, then $0!$ also doesn't make sense similar to that of negative integers right? – Rayreware Sep 07 '19 at 05:22
  • @Rayreware See https://en.wikipedia.org/wiki/Empty_product – Vsotvep Sep 07 '19 at 05:24
  • @Vsotvep, yes agreed $0!=1$, and by that same logic why can't we establish $0^0=1$ too because $0^0$ returns a null set right? – Rayreware Sep 07 '19 at 05:39
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    @Rayreware Please see the other questions for what I would answer. It depends on context. I suggest to stop this discussion here, since it has been done many times elsewhere already. – Vsotvep Sep 07 '19 at 05:54
  • https://math.stackexchange.com/questions/3198313/a-thorough-explanation-on-why-division-by-zero-is-undefined/3198337#3198337 –  Sep 08 '19 at 21:51

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