So I thought about this for a little longer, and now I actually think that Stefan's Formula is true for all limit ordinals $\delta$ and strictly increasing sequences of cardinals.
Lemma 1: Let $\kappa_1≤\kappa_2$, $\lambda_1≥\lambda_2$ be infinite cardinals such that $cof(\kappa_2)≤\lambda_2$. Then $\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}=\kappa_2^{\lambda_2}$.
Proof: Obviously we have $\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}≤\kappa_2^{\lambda_1}$. We fix $\kappa_1, \lambda_1, \lambda_2$ and show "≥" by induction on $\kappa_2$ starting with the trivial case $\kappa_2=\kappa_1$.
Let's distinguish 3 cases:
case 1: $\lambda_1≥\kappa_2$: $\kappa_2^{\lambda_1}≤\lambda_1^{\lambda_1}=2^{\lambda_1}≤\kappa_1^{\lambda_1}≤\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}$
case 2: $\lambda_1<\kappa_2$ and $\forall\xi<\kappa_2:\xi^{\lambda_1}<\kappa_2$. It is known that in this case $\kappa_2^{\lambda_1}$ is either $\kappa_2$ (if $cof(\kappa_2)>\lambda_1$) or $\gimel(\kappa_2)$ (else). In both cases it is smaller (EDIT: or equal) than $\kappa_2^{\lambda_2}$
case 3: There exists some $\xi<\kappa_2$ such that $\xi^{\lambda_1}≥\kappa_2$. Then by induction $\kappa_2^{\lambda_1}≤\left(\xi^{\lambda_1}\right)^{\lambda_1}=\xi^{\lambda_1}≤\kappa_1^{\lambda_1}\xi^{\lambda_2}≤\kappa_1^{\lambda_1}\kappa_2^{\lambda_2}$
Lemma 2: Let $\kappa_0, \kappa_l$ and $\lambda_0, \lambda_l$ be finitely many infinite cardinals such that $\forall k≤ l: cof(\kappa_k)≤\lambda_k$. Then $\max\limits_{0≤k≤ l}\kappa_k^{\lambda_k}=\prod\limits_{k=0}^l\kappa_k^{\lambda_k}=\max\limits_{0≤k,j≤ l}\kappa_k^{\lambda_j}=\left(\max\limits_{k≤ l}\kappa_k\right)^{\max\limits_{k≤ l}\lambda_k}$.
Proof: This follows immediately from Lemma 1 by induction.
Now let $\delta$ be a limit ordinal and $(\kappa(i)\mid i<\delta)$ a nondecreasing sequence of ordinals. Then writing $\delta=\lambda_0\alpha_0+\ldots+\lambda_l\alpha_l$ and $\delta_k:=\lambda_0\alpha_0+\ldots+\lambda_k\alpha_k$ as above (with $0<\alpha_k<\lambda_k^+$) we observe that all $\lambda_k≥\omega$ (since $\delta$ limit).
Then writing $\mu_k:=\sup\limits_{i<\lambda_k\alpha_k}\kappa(\delta_k+i)$ we saw $\prod\limits_{i<\delta}\kappa(i)=\max\limits_{k≤ l}\mu_k^{\lambda_k}$.
But notice that by definition $(\kappa(\delta_k+i)\mid i<\lambda_k\alpha_k)$ is cofinal in $\mu_k$, so $cof(\mu_k)≤|\lambda_k\alpha_k|=\lambda_k$. This by Lemma 2 gives us $\max\limits_{k≤ l}\mu_k^{\lambda_k}=\left(\max\limits_{k≤ l}\kappa_k\right)^{\max\limits_{k≤ l}\lambda_k}=\mu_l^{\lambda_0}=\left(\sup\limits_{i<\delta}\kappa(i)\right)^{|\delta|}$