Suppose that we start with some prime $p$. We then take a square of $p$ and sum that square with some number from the set $\{1,2,...p\}$ to obtain some other prime. We then cube $p$ and sum that cube with some number from the set $\{1,2,...p^2\}$ to obtain some prime again. At $n$-th step we take $p$ to the power of $n$ and sum $p^n$ with some number from the set $\{1,2,..p^{n-1}\}$ to obtain some prime. We do that until it is possible to do that.
An example: $p=7$. Squaring gives $7^2=49$ we now must sum $49$ with some number from the set $\{1,2,...7\}$ to obtain a prime. We can choose $4$ because $49+4=53$, a prime. We then cube $7$ to obtain $7^3=343$ and now we choose some number from the set $\{1,2,...49\}$ and we can choose $4$ to obtain $343+4=347$. And so on...
Is there a prime $p$ such that this procedure fails after a finite number of steps?
A question, phrased differently is:
Is there for every prime $p$ a prime in the set $\{p^n+1,...p^n+p^{n-1}\}$ for every $n \in \mathbb N \setminus \{2\}$?
Assuming the highly likely truth of Oppermann's conjecture, there will be a prime between $n^2$ and $n^2+n$. The slightly shaky region is up to about $200,$ after which gaps between primes are comfortably smaller than the square root of the relevant primes.
Actual prime gaps become relatively far smaller as $n$ gets big - the largest numerical prime gap among the fully identified prime set is only $1510$, at $p\approx 6.8\times 10^{18}$
As Robert Israel mentioned, this (first step) is an open issue, Oppermann's conjecture as stated. The difficulty of proving the bounds of such prime gaps is out of step with the size of gaps themselves, the largest of which are far smaller than the known bounds.