Let $Card$ be the class of infinite cardinals and $p\colon Card^2\to Card$ be given by $(\kappa,\lambda)\mapsto\kappa^\lambda$.  

Assuming GCH it is known that $p(\kappa,\lambda)$ is either $\kappa$ (if $\lambda < cof(\kappa)$) or $\kappa^+$ (if $cof(\kappa)≤\lambda ≤\kappa$) or $\lambda^+$ (if $\kappa < \lambda$).  

However, knowing less can be enough.
For instance the function $\gimel\colon \kappa\mapsto\kappa^{cof(\kappa)}$ completely determines $p$. 


Generalizing from powers to arbitrary products yields the following natural question:


Consider a sequence $(\kappa\_i\mid i<\delta)$ of (infinite) cardinals. Can the value $\prod\limits\_{i<\delta}\kappa\_i$ be de determined knowing (all values of) $\gimel$ or assuming GCH?


One can reformulate the question as: under which circumstances is $\prod\limits\_{i<\delta}\kappa\_i$ determined by $p$?


EDIT: So I guess this is the final result:  

For $(\kappa(i)\mid i<\delta)$ nondecreasing sequence of infinite cardinals we have for $\delta=\lambda\_0\alpha\_0+\ldots+\lambda\_l\alpha\_l$ (every ordinal can be written this way) and $\delta\_k:=\lambda\_0\alpha\_0+\ldots+\lambda\_{k}\alpha\_{k}$ with cardinals $\lambda\_0>\ldots>\lambda\_l$ and ordinals $0<\alpha\_k<\lambda\_k^+$:  

$\prod\limits\_{i<\delta}\kappa(i) = \max\limits\_{0 ≤ k ≤ l} \left(\sup\limits\_{i<\lambda\_k\alpha\_k}\kappa(\delta\_{k-1}+i)\right)^{\lambda\_k}$



do behavior of gimel or GCH determine all infinte products of cardinals?
I guess this is the final result:  

For $(\kappa(i)\mid i<\delta)$ nondecreasing sequence of infinite cardinals we have for $\delta=\lambda\_0\alpha\_0+\ldots+\lambda\_l\alpha\_l$ (every ordinal can be written this way) and $\delta\_k:=\lambda\_0\alpha\_0+\ldots+\lambda\_{k}\alpha\_{k}$ with cardinals $\lambda\_0>\ldots>\lambda\_l$ and ordinals $0<\alpha\_k<\lambda\_k^+$:  

$\prod\limits\_{i<\delta}\kappa(i) = \max\limits\_{0 ≤ k ≤ l} \left(\sup\limits\_{i<\lambda\_k\alpha\_k}\kappa(\delta\_{k-1}+i)\right)^{\lambda\_k}$




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Lemma 1: Each limit ordinal $\alpha$ can be written as $\sum\limits\_{i< l}\lambda\_i\alpha\_i$ where $(\lambda\_i\mid i< l)$ is a strictly decreasing finite sequence of infinite cardinals and $(\alpha\_i|i< l)$ are some ordinals with $\alpha\_i<\lambda\_i^+$.


Proof: Take some $\alpha\in Lim$. Let $\lambda\_0:=|\alpha|$ and $\beta:=min(\gamma\in Ord\mid \lambda \cdot \gamma >\alpha)$. If $\beta$ was a limit we would have $\lambda\beta = \sup\limits\_{\gamma<\beta}\lambda\gamma≤\alpha$, so $\beta$ is a successor and we can write $\beta=\alpha\_0 +1$.  

Clearly $|\alpha\_0|<\lambda^+$ since $|\lambda\alpha\_0|≤|\alpha|=\lambda$.  

Now $\alpha-\alpha\_0$ has cardinality smaller than $\lambda$ (else $\lambda\alpha\_0+\lambda=\lambda\cdot\beta≤\alpha$), so we are done by induction on the cardinality.




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Lemma 2: Let $\lambda$ some infinite cardinal, $0<\alpha<\lambda^+$ some ordinal and $X$ some set of cardinality $\lambda$. Then there is a bijection $t=(t\_1,t\_2)\colon\lambda\alpha\to\lambda\alpha\times X$ such that $t\_1≤id\_{\lambda\alpha}$.


Proof: Induction on $\alpha$. For $\alpha=1$ identify $X\sim\lambda$ and take the Mostowski collapse of the Cantor wellordering on $\lambda \times\lambda$ (given by ordering pairs first by their maximum, then by the first and then by the second component). It is known that this is a bijection $t\colon\lambda\times\lambda\leftrightarrow\lambda$ and $t\_1≤id$ is clear.  

Else write each $\beta<\alpha$ as $\lambda\gamma+\epsilon$ with $\gamma$ maximal (like in proof of Lemma 1.) and map it to $(\lambda\gamma + t\_1(\epsilon),t\_2(\epsilon))$.




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Lemma 3: Let $\lambda$ some infinite cardinal, $0<\alpha<\lambda^+$ some ordinal. For every nondecresing sequence $(\kappa\_i\mid i<\lambda\alpha)$ we have $\prod\limits\_{i<\lambda\alpha}\kappa\_i≥\left(\prod\limits\_{i<\lambda\alpha}\kappa\_i\right)^{\lambda\alpha}$.


Proof: We have an injection  

$A:=${$f:\lambda\alpha\times\lambda\alpha\to V\mid\forall i,j <\lambda\alpha: f(i,j)\in \kappa\_i$} $\to$ {$g\colon\lambda\alpha\to V\mid\forall i<\lambda\alpha: g(i)\in\kappa$}$=:B$ given by $f\mapsto f\circ t$, where $t$ is as in Lemma 2 for $X:=\lambda\alpha$.  

But $\prod\limits\_{i<\lambda\alpha}\kappa\_i = |B|$, $\left(\prod\limits\_{i<\lambda\alpha}\kappa\_i\right)^{\lambda\alpha}=|A|$.




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If we have a sequence of cardinals $(\kappa\_i|i<\delta)$ we can determine the product as follows: 


Write $\delta=\sum\limits\_{k≤ l}\lambda\_k\alpha\_k$ as in Lemma 1. Determine the product of the first $\delta':=\sum\limits\_{k≤ l-1}\lambda\_k\alpha\_k<\delta$ factors by induction, if $0< l$. So all that's left to do is find out what $\prod\limits\_{i<\lambda\_l\alpha\_l}\kappa\_{\delta'+i}$ is:


As before we can assume that $\delta$ is a limit, so $\lambda\_l\alpha\_l$ is too.
Then set $\kappa=\sup\limits\_{i<\lambda\_l\alpha\_l}\kappa\_{\delta'+i}$ and then by Lemma 3:
$\kappa^{\lambda\_l}=\kappa^{\lambda\_l\alpha\_l}≤\left(\prod\limits\_{i<\lambda\_l\alpha\_l}\kappa\_{\delta'+i}\right)^{\lambda\_l\alpha\_l}≤\prod\limits\_{i<\lambda\_l\alpha\_l}\kappa\_{\delta'+i}≤\kappa^{\lambda\_l\alpha\_l}=\kappa^{\lambda\_l}$


