By 1 step breach of the GCH I mean the following: $$ 2^{\aleph_{\alpha}} = \aleph_{\alpha+2}$$
Now, it is known that there are more constrains on the cardinality of power sets at singlular cardinals that their cardinality is influenced by the behaviour of the continuum below them! The main theorem supporting that is Silver's Theorem which states that: if GCH holds below a singular of uncountable cofinality, then it holds at that singular as well.
My question is about if a similar result to Silver's theorem is true in $\sf ZFC$, that is if the GCH is breached in the same manner before a singular of uncountable cofinality, then that breach holds at that singular as well! Formally, that is: $$ \operatorname {singular}(\aleph_\lambda) \land \\ \operatorname {cf}(\aleph_\lambda) > \omega \land \\ \forall \alpha < \lambda \, (2^{\aleph_\alpha}= \aleph_{\alpha+2}) \\\to \\ 2^{\aleph_\lambda}= \aleph_{\lambda+2}$$
In English: if there is 1 step breach of the GCH below a singular of uncountable cofinality, then there is a 1 step breach of GCH at that singular itself.
Of course the more general theorem would be about $n$-step breach of GCH for a finite $n$, or it might be possible to even extend it to some infinite values?
Best Answer
One can collapse $2^\lambda$ to have cardinality $\lambda^+$ without adding $\lambda$-sequences even if $\lambda$ is singular. Therefore one could start with $2^{\aleph_\alpha} = \aleph_{\alpha+2}$ for all $\alpha < \omega_1$ by Foreman-Woodin and then force $2^{\aleph_{\omega_1}} = \aleph_{\omega_1+1}$ without changing $P(\aleph_{\omega_1})$ and hence preserving $2^{\aleph_\alpha} = \aleph_{\alpha+2}$ for all $\alpha < \omega_1$. This shows that the answer to your question is no.