Is it consistent with ZFC that for all ordinals $\alpha, \beta < \omega$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$? Let $\gamma=\omega$ (the first transfinite ordinal). Is it consistent with ZFC that for all ordinals $\alpha, \beta < \gamma$ it holds that $2^{\aleph_\alpha} = 2^{\aleph_\beta}$?
If yes, can the bound $\gamma$ be increased here and how much?

Update: In what sense the bound $\gamma$ can be made arbitrarily high? If $\beta$ is the initial ordinal of $\beth_1$, then it cannot be that $2^{\aleph_0}=2^{\aleph_\beta}$, right?
 A: Yes.  Start with a model of GCH and add $\aleph_{\omega+1}$ Cohen reals.  Then $2^{\aleph_n}=\aleph_{\omega+1}$ for all $n<\omega$.  You can get the bound $\gamma$ arbitrarily high within the ordinal hierarchy by adding $\kappa$ Cohen reals instead, where $\kappa$ is a regular cardinal greater than $\aleph_\gamma$.  (I think that's all correct.)
A: Yes. A fairly general answer to such questions is provided by Easton’s theorem: if the ground model satisfies GCH and $F$ is a class function from a subclass of regular cardinals to cardinals such that $F$ is nondecreasing and $\kappa< \operatorname{cf}(F(\kappa))$ for each $\kappa\in\operatorname{dom}(F)$, then one can construct a forcing extension with the same cardinals and cofinalities which satisfies $2^\kappa=F(\kappa)$ for each $\kappa\in\operatorname{dom}(F)$.
A: To your edit, note that for every ordinal $\alpha$ it holds that $\alpha\le\aleph_\alpha$. This is because there are $\alpha$ many cardinals below $\aleph_\alpha$.
Since the function $\kappa\mapsto 2^\kappa$ is strictly increasing, we have if so that $2^{\aleph_0}=\beta<2^{\beta}\le 2^{\aleph_\beta}$.
