Question about zero times infinity (distinct from previous) 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 $0\times\infty$.

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 $0\times \infty = 0$.
