Do $p^n$-th powers determine the field? This is a question which came to my mind, when seeing
A quick question on transcendence
Suppose $F$ is a field of characteristic $p$. Then the set of $p^n$-powers of the elements
of $F$ is again a field which I denote by $F^n$ and is a subfield of $F$.
Now let $d_n=[F:F^n]$.
My question is, what can be said about $F$ if we know the sequence $(d_n)_n$.
Of course for finite fields $d_n$ is monotonically increasing and eventually constant.
The constant will be the index degree of $F$ over the prime field $F_p$ and thus
$F$ is determined up to isomorphism.
So what happens in general. Of course $d_n$ is monotonically increasing. In general $d_n$ will not be finite. This can be seen by taking $F=F_p(t_1,t_2,\ldots)$ where $t_i$ is transcendental over $F_p(t_1,t_2,\ldots,t_{i-1})$.
So now my questions:


*

*Can it happen that $d_1$ is finite but $d_n$ is infinite for some $n$?

*Under the assumption that $(d_n)_n$ is finite and the same for a field $F_1$ and $F_2$.  Are these fields necessarily isomorphic?
 A: First, your notation seems potentially misleading: it is standard to use $F^n$ to denote the set of $n$th powers of elements of $F$.  So I will use $F^{p^n}$ instead.
Second, your assertions about finite fields are incorrect (perhaps you got confused by your notation, as above?).  In fact for any perfect field we have $d_n = 1$ for all $n$.
Here is a key observation: if $E/F$ is a field extension and $\sigma: E \rightarrow E$ 
is a field homomorphism, then $[E:F] = [\sigma(E):\sigma(F)]$.  Indeed, if $\{x_i\}$ 
is a basis for $E/F$, then $\{\sigma(x_i)\}$ is abasis for $\sigma(E)/\sigma(F)$.  Applying this with $F/F^p$ and $\sigma = (x \mapsto x^p)$, we get $[F:F^p] = [F^p:F^{p^2}]$ and thus 
$d_2 = [F:F^{p^2}] = [F:F^{p}] [F^{p}:F^{p^2}] = [F:F^p] [F:F^p] = d_1^2$.
Reasoning in a similar way gives $d_n = d_1^n$ for all $n \in \mathbb{Z}^+$.  So you are not getting any new information from the entire sequence $\{d_n\}$ that you don't already know from $d_1$.
As one might suspect, there are lots of fields having the same value of $d_1$.  For instance, as above, all perfect fields have $d_1 = 1$.  Moreover, for any positive integer $d$, any two function fields of transcendence degree $d$ over any algebraically closed field of characteristic $p$ will have $d_1 = p^d$. 
