Is there a $c > 1$ such that for all $n \ge 1$ the largest integer $\le c^n$ is prime? Does there exist a real number $c > 1$ such that for every natural number $n > 0$, the number $\lfloor c^n \rfloor$ is prime?
I doubt such a number $c$ is known to exist, since the best similar results I've seen are much weaker.  For example, in 1947  William Harold Mills proved that there is a real number $c > 1$ such that for every natural number $n > 0$, the number $\lfloor c^{3^n} \rfloor$ is prime:

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*Wikipedia, Mills' constant
and if the Riemann Hypothesis holds, the smallest such $c$ is approximately
$$ c = 1.3063778838630806904686144926\ldots $$
Harder results along these lines are known, e.g. in 2010 Matomäki showed that there exists an uncountable infinity of real numbers $c > 1$ with the property that for every natural $n > 0$, the number $\lfloor c^{2^n} \rfloor$ is prime:

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*K. Matomäki, Prime-representing functions, Acta Mathematica Hungarica 128 (4) (2010), 307–314.

However, I haven't seen results like this for $\lfloor c^n \rfloor$.  So, I think my question boils down to: is the answer to my original question known to be no, or is it still open?
 A: I rewrote Sebastien Palcoux's answer in Mathematica as f[{x_, n_}] := If[PrimeQ[Floor[x^n]], {x, n+1}, {NextPrime[x^n]^(1/n), 1}].
Iterating this ten times with NestList[f, {2, 1}, 10] gives, as in his answer:
$$\{2,1\},\{2,2\},\\
\left\{\sqrt{5},1\right\},
\left\{\sqrt{5},2\right\},\left\{\sqrt{5},3\right\},
\left\{\sqrt{5},4\right\},\\\left\{29^{1/4},1\right\},
\left\{29^{1/4},2\right\},\left\{29^{1/4},3\right\},\\
\left\{13^{1/3},1\right\},\dots$$
Iterating this 1,047,399 times confirms that any $c$ would satisfy $c\ge 40$.
Iterating this three million times, with about four minutes of computation time, finds a highest exponent of 27: it produces three values of $x$ with $41\le x\le 42$ where the floors of $x^1,\ldots,x^{27}$ are all prime. In fact $x^{27}$ is a 44-digit prime in all three of those cases.
Other smaller but interesting cases include $x=7691^{1/3}$ and $x=13591^{1/3}$, for which the first 5 powers have prime floors; and $x=32340221^{1/5}$, for which the first 8 powers have prime floors.
A: The question is equivalent to that of asking whether the sets
$$ E_n := \bigcup_p [\frac{\log p}{n}, \frac{\log(p+1)}{n})$$
for $n=1,2,\dots$ have a non-empty intersection, where $p$ ranges over primes (since $\log c$ would have to lie in all of the $E_n$ in order for the claimed property of $c$ to hold).  Probabilistic heuristics suggest that the answer is negative, though a proof seems well out of reach of current technology.  Heuristically, it suffices to show that for any $1 < a < b$, the sets $E_n \cap [a,b]$ have empty intersection "almost surely".  In particular, it would heuristically suffice to show that as $N \to \infty$, the "probability" that
$$ \bigcup_{N \leq n < 2N} E_n \cap [a,b] = \emptyset$$
goes to zero as $N \to \infty$.
Observe that if $\frac{k}{b^{3N}} \in [a,b]$ lies within $\frac{1}{b^{3N}}$ of $E_n$, then $\frac{k}{b^{3N}} = \frac{\log p}{n} + O( \frac{1}{pn})$ for some prime $p \in [a^n,b^n]$.  A calculation using the prime number theorem then shows that the discretized set
$$ F_n := \{\frac{k}{b^{3N}} \in [a,b]: k \in {\bf N}, \mathrm{dist}(\frac{k}{b^{3N}}, E_n) \leq \frac{1}{b^{3N}} \}$$
has density $O(1/N)$ in
$$ F_0 := \{\frac{k}{b^{3N}} \in [a,b]: k \in {\bf N} \}$$
(we allow implied constants to depend on $a,b$). Since $F_0$ is a finite set of cardinality $O( b^{3N} )$, and there is no reason to expect the sets $F_n$ to ``correlate'' with each other, probabilistic heuristics suggest that $\bigcap_{N \leq n < 2N} F_n$ should be empty with "probability" $O(b^{3N}) (O(1/N))^N$, which goes to zero as $N \to \infty$, and hence the "probability" that $\bigcap_{N \leq n < 2N} E_n \cap [a,b]$ is empty will do so also.
