Cardinality of power sets decides all of cardinal arithmetic? Assuming ZFC, is it possible to have two models which agree on the cardinality of all the power sets, but disagree on the cardinality of some other cardinal exponentiation (meaning that they agree on the function $F$ such that $F(\alpha) = \beta$ iff $2^{\aleph_\alpha} = \aleph_\beta$, but in one model ${\aleph_\alpha} ^ {\aleph_\beta} = {\aleph_\gamma}$ whereas in the second model ${\aleph_\alpha} ^ {\aleph_\beta} = {\aleph_\delta}$ with $\gamma \neq \delta$)?
Put another way: If we decide (for example by forcing) the cardinality of every power set ($2^{\aleph_\alpha}$ for all $\alpha$), does it automatically decide the result of every possible cardinal exponentiation (${\aleph_\alpha} ^ {\aleph_\beta}$ for all $\alpha,\beta$)?
For example, when we assume GCH, we have an immediate formula for the cardinality of every cardinal exponentiation (see Jech 5.15), and we have no "freedom" to choose alternative values for them.
 A: The answer is no assuming the existence of a model of $\mathsf{ZFC+GCH}$ having a supercompact cardinal.
First, using Silver's forcing, there is a generic extension $V[K]$ where $\kappa$ is still measurable but $2^{\kappa}=\kappa^{++}$. Then we can use Prikry's forcing to obtain a generic extension $V[K][H]$ of $V[K]$ such that all bounded subsets of $\kappa$ are in $V[K]$, all cardinals are preserved, $\kappa$ is still a strong limit and $\operatorname{cf}\kappa=\omega$. Let $G=K\ast H$.
As we assumed $V\models\mathsf{GCH}$, it's not hard to prove that in $V[G]$ we have $2^\lambda=\lambda^+$  for all $\lambda>\kappa$; since the poset yielding $V[K]$ has size $\kappa^{++}$ in $V$ and the poset giving the extension $V[K][H]$ has size $\kappa^{++}$ in $V[K]$. 
Now let us work in $V[G]$.  We have $$\kappa^{\aleph_0}=\kappa^{\operatorname{cf}\kappa}=2^{\kappa}=\kappa^{++},$$ and $$(\kappa^{+3})^{\omega_1}=(2^{\kappa^{++}})^{\omega_1}=2^{\kappa^{++}}=\kappa^{+3},$$ thus we can force  with $Add(\omega_1,\kappa^{+3})$ to obtain a generic extension $V[G][H']$ where $2^{\omega_1}=\kappa^{+3}$, and $2^{\lambda}=\kappa^{+3}$ for all $\omega_1\leq\lambda\leq \kappa^{++}$. In $V[K]$, $\kappa$ is measurable, thus in there  $\kappa=\aleph_\kappa$, so as cardinals are preserved in $V[G]$ we get that $\kappa=\aleph_\kappa$ is also true in $V[G][H']$. 
Let $\beta_0$ be such that $2^{\aleph_0}=\aleph_{\beta_0}$ in $V[G]$. Then $\beta_0<\kappa$; as all bounded subsets of $\kappa$ in $V[G]$ are in $V[K]$. We also have $2^{\aleph_0}=\aleph_{\beta_0}$ in $V[G][H']$.
Thus if we consider the following function
$$F(\alpha)=\begin{cases} \beta_0 & \text{if}&\alpha=0 \\\kappa+3 & \text{if}& 1\leq\alpha\leq\kappa+2\\\alpha+1 &\text{if}&\alpha\geq\kappa+3 \end{cases},$$
it follows that for all ordinals $\alpha$, $$V[G][H']\models 2^{\aleph_\alpha}=\aleph_{F(\alpha)},$$
and as the poset we used in $V[G]$ is $<\omega_1$-closed there, we get that $\kappa^{\aleph_0}=\kappa^{++}$ in this extension too.
Now, let $\mathbb P\in L$ be a poset, cardinal preserving, such that if $K'$ is $L$-generic over $\mathbb P$, we have  for all ordinals $\alpha$, $$L[K']\models 2^{\aleph_\alpha}=\aleph_{F(\alpha)}.$$ 
In $L$, $\kappa$ is inaccesible, and thus in this model $\aleph_\kappa=\kappa$, so as $\mathbb P$ preserves cardinals we get that $\aleph_\kappa=\kappa$ in $L[K']$ too. 
Let us work in $L[K']$. The singular cardinals hypothesis is true; since $0^\sharp$ does not exist, so we get that as $2^{\aleph_0}<\kappa$ and $\kappa$ is regular, $\kappa^{\aleph_0}=\kappa$.
Therefore we have that in both models $V[G][H']$ and $L[K']$, $2^{\aleph_\alpha}=\aleph_{F(\alpha)}$ for all ordinals $\alpha$, but
$$V[G][H']\models \aleph_\kappa^{\aleph_0}=\aleph_{\kappa+2}\text{ and }L[K']\models \aleph_\kappa^{\aleph_0}=\aleph_\kappa.$$

Note: This argument should go through with no problem using just a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}$, working in Mitchell's model for such $\kappa$, using Gitik and Woodin's forcing. However as I'm not that familiar with this method, I used Silver's instead. 
