Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integers $n$? Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integer $n$?

In general, let $x_{k}$ denote the base-$k$ representation of the positive integer $x$. We say $x$ is $k$-powerful if $x^n_{k}$ contains all of the $k$ digits for all sufficiently large integers $n$.
For example, it's easy to check that 2 is 2-powerful, but 3 is not 3-powerful. The title question asks whether 2 is 10-powerful.
For any given $x$ and $k$, can we decide if $x$ is $k$-powerful?
 A: Heuristically, one would expect the answer to be yes. There's an existing partially explicit version of this mentioned in Richard Guy's "Unsolved Problem in Number Theory" entry F24, which is that for $n> 86$, $2^n$ always contains a zero in its base 10 expansion. There are some related questions also in that entry and some other entries in the book as well. For example, there's a conjecture that every sufficiently large power of 2 contains 0s, 1s and 2s in its base 3 expansion. The reasonable generalization (which I have not seen stated explicitly but seems to be implicit in all of these), is that if $a$ and $b$ are relatively prime integers both greater than 1, then for all sufficiently large $n$, $a^n$ has in its base $b$ expansion all of $0, 1, \dotsc b-1$. It is also plausible that the sufficiently large is a not very fast growing function, something like being true for all $n \geq a^2b^2$. Edit: See Emil's comment below: as written this doesn't include the case $2$ and $10$; one wants that there's a prime $p$ such that $p$ divides exactly one of $a$ and $b$. That includes the relatively prime case and also cases like $2$ and $10$.
However, the set of $a$ and $b$ where we can prove anything like this is small and mostly relegated to when $b=2$. In that case, some of these results are implicitly very old, dating back to actually the middle ages. For example, any power of $3$ greater than $3$ must contain both $1$ and $0$ in its base $2$ expansion. This follows, since if not, $3^n$ would have to be of the form $2^k-1$, and Gersonide's theorem that $8$ and $9$ are the largest power of 2 next to a power of $3$ then applies. For larger bases, Mihailescu's proof of Catalan's conjecture shows that the same is essentially true if one has $b=2$ and $a$ is any number greater than $2$.
