I know there are very many threads about this question already and I have perused them to make sure this is unique.

The quantity $0^\infty$ perfectly well defined. Therefore, when we have the expression $0\times\infty$, what error is invoked when I do the manipulation

$$ 0\times\infty= 0^\infty\times 2^\infty=(0\times2)^\infty=0^\infty=0~~?$$

The above is one case where I have conjured $2$ from "the ether," so let me give another example where I don't have to "deconstruct" infinity as I have with $\infty=2^\infty$. Let's say I have an infinite product function of an extended real number (reals including infinity) so that the domain is $x\in[-\infty,\infty]$:

$$ f(x) = \prod_{k=1}^\infty \dfrac{(1+k)^x}{x} $$

It follows that

$$ f(\infty) = \prod_{k=1}^\infty \dfrac{(1+k)^\infty}{\infty}= \prod_{k=1}^\infty \bigg[0\times (1+k)^\infty \bigg] $$

Since the least value of $k$ is $1$, every term $(1+k)^\infty$ is going to be equal to infinity and, if I evaluate it, I will get $f(\infty) = \Pi (0\times\infty)=$ undefined. However, before I evaluate the $k$ term, I can use $0=0^\infty$, and then write

$$ f(\infty) = \prod_{k=1}^\infty \bigg[0^\infty\times (1+k)^\infty \bigg] = \prod_{k=1}^\infty \bigg[ [0\times (1+k)]^\infty \bigg]=0 $$

Since there is a way for me to do the operations which does not lead to an undefined expression, is this order of operations preferred. Any insight much appreciated.

EDIT: This question differs from the one it is alleged to duplicated because it asks questions about exponent operations that do not appear in the alleged duplicate question. Furthermore, the question about the structure of the infinite product is completely disconnected from the alleged duplicate question, other than it does it relate to the same parent issue $0\times\infty$.

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    $\begingroup$ "The quantity $0^∞$ perfectly well defined" ? $\endgroup$ – Mauro ALLEGRANZA Jun 27 '19 at 11:35
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    $\begingroup$ The issue is : are we entitled to treat $\infty$ as a real number ? If so, $0 \times n=0$ for every $n$. If not, $0 \times \infty$ is not defined (at least when we use the "usual" def of product between real numbers). $\endgroup$ – Mauro ALLEGRANZA Jun 27 '19 at 11:38
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    $\begingroup$ Infinity is not a number. No calculation rules are valid when dealing with infinity (certainly not $a^{\infty}b^{\infty}=(ab)^{\infty}$). $\endgroup$ – Mindlack Jun 27 '19 at 11:39
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    $\begingroup$ @Mindlack Infinity is not a real number. There are systems of numbers that include transfinites. $\endgroup$ – Sambo Jun 27 '19 at 11:44
  • $\begingroup$ @Sambo: You are right. However, I don’t know of any transfinite system using infinity as such. You write it as $\aleph$, $\omega$, $\lambda$, and so on... but never simply $\infty$ (to my knowledge). $\endgroup$ – Mindlack Jun 27 '19 at 11:50

Here is a contradiction if we are allowed to manipulate infinity in the exponents as you claim:

\begin{align} \infty &= 2^{\infty} \\ &= ((1/2)(4))^{\infty} \\ &= (1/2)^{\infty} 4^{\infty}\\ &= 0 \times \infty \end{align} which contradicts your claim that $0\times \infty = 0$.

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  • $\begingroup$ This uses the basic definition: If $a\geq 0$ then $a^{\infty} = \lim_{n\rightarrow\infty} a^n$. (This definition seems to be implied in the question and is consistent with $0^{\infty}=0$ and $(1+k)^{\infty} = \infty$ for all $k> 0$.) The above argument shows that, even with such a definition, we are not allowed to manipulate $\infty$ in the exponents in the same way that real numbers can be manipulated. $\endgroup$ – Michael Jun 27 '19 at 19:30
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    $\begingroup$ Nice! Thanks for this clear statement. You could have chosen to only hint at this contradiction but, instead, you clearly stated it. Users like you make this a better website. $\endgroup$ – hodop smith Jun 27 '19 at 21:37

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