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 $0<\alpha_k<\lambda_k^+$:

$\prod\limits_{i<\delta}\kappa(i) = \max\limits_{0 ≤ k ≤ l} \left(\sup\limits_{i<\lambda_k\alpha_k}\kappa(\delta_{k-1}+i)\right)^{\lambda_k}$

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.

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 $(\lambda\gamma + t_1(\epsilon),t_2(\epsilon))$.

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 $\prod\limits_{i<\lambda\alpha}\kappa_i = |B|$, $\left(\prod\limits_{i<\lambda\alpha}\kappa_i\right)^{\lambda\alpha}=|A|$.

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:
$\kappa^{\lambda_l}=\kappa^{\lambda_l\alpha_l}≤\left(\prod\limits_{i<\lambda_l\alpha_l}\kappa_{\delta'+i}\right)^{\lambda_l\alpha_l}≤\prod\limits_{i<\lambda_l\alpha_l}\kappa_{\delta'+i}≤\kappa^{\lambda_l\alpha_l}=\kappa^{\lambda_l}$