Is this behavior of the continuum function consistent with ZFC. I have an axiom, based on how finite cardinalities work. As we know, every ordinal can be written as the sum of a unique limit ordinal $L$ (where $0$ is a limit ordinal axiom for the purposes of this axiom) and a unique natural number $n$. My axiom is that $2^{\aleph_{L+n}}=\aleph_{L + 2^n}$, where $\aleph$ is Cantor's $\aleph$ cardinality function. 
My question is, is this axiom consistent with ZFC?
 A: Yes. This is consistent.
Recall Easton's theorem, that if we begin with a model of $\sf ZFC+GCH$ and $F$ is a function defined on regular cardinals, such that $\alpha\leq\beta\implies F(\alpha)\leq F(\beta)$ and $\operatorname{cf}(\alpha)<\operatorname{cf}(F(\alpha))$, then there is a [class-]generic extension of the universe such that $2^\alpha=F(\alpha)$.
As luck would have, if $\alpha$ is a regular cardinal, say $\aleph_{\delta+n}$ where $\delta$ is limit ($n$ may or may not be $0$), then $F(\aleph_{\delta+n})=\aleph_{\delta+2^n}$ satisfies the necessary conditions indeed. So we can get that for regular cardinals.
But not all cardinals are regular. What about singular cardinals? Easton's theorem gives us a model where for a singular cardinal $\lambda$, $2^\lambda$ is the least possible value. In the case of $F$ as we have here, since we start in a model of $\sf GCH$, the result is a model where limit cardinals remain strong limit cardinals. So the least possible value is $\lambda^+$. But as luck would have it, if $\lambda$ is singular, then for some limit ordinal $\delta$, $\lambda=\aleph_{\delta+0}$ and $\lambda^+=\aleph_{\delta+1}=\aleph_{\delta+2^0}$.
So we get a model as wanted.
