Is there a $k$ such that $2^n$ has $6$ as one of its digits for all $n\ge k$?

Question

It is true that every power of $2$ of the form $2^{6+10x}$, $x\in\mathbb{N}$, has $6$ as one of its digits. Something more is true, the last two digits are either $64$ or $36$. The OP suggests that "...Beyond that the numbers are so large that they seem to all contain a 6 as part of their digits." This leads to the following related questions.

Q.1. Is there a natural number $k$ such that $2^n$ has $6$ as one of its digits for all natural numbers $n\ge k$ ?

Q.2. Is there a natural number $k$ and a natural number $j$ such that the last $j$ many digits of $2^n$, for $n\ge k$, would cycle through only finitely many possibilities, each of which has a $6$ as one of its digits? I believe that formally this is asking the following. Are there triples $(k,j,\mathcal F)$ where $k,j$ are natural numbers and $\mathcal F$ is a finite set of numbers each with $j$ many digits one of which is necessarily $6$ (leading zeroes allowed) such that $2^n\mod 10^j \in\mathcal F$ whenever $n\ge k$ ? ("Cycling" also suggests that there is a particular order under which the elements of $\mathcal F$ appear, as $n$ increases, and this order repeats, that is say $\mathcal F=\{F_0,F_1,...,F_{m-1}\}$ and $2^n\mod 10^j =F_{(n \mod m)}$ whenever $n\ge k$, where each $F_t$ has $6$ as one of its digits. So there are two versions of Q.2, with or without cycling. Here $2^n\mod 10^j$ is of course understood as a function, $\mod(2^n,10^j)$, with values in the set $\{0,1,...,10^j-1\}$.)

Q.3. Which is, on its own, several more questions, namely the variations of Q.1 and Q.2 when we use a different digit instead of $6$? (And even more variations, what if we consider a different base than $2$?)

To keep it focused, Q.1 is the "official" question posted here, but clearly Q.2 is very closely related (and perhaps even more interesting), and Q.3 seems related too (but feels too broad for now, perhaps should be posted separately). (Of course having this question depend on decimal representation makes is a bit lacking in motivation, but it might be interesting anyway.)

Edit. Collecting some of the comments posted during the first two hours, plus some numerical data.

I agree with @LaBird that the answer to Q.1 is likely yes, with $k=94$. There is no 6 in $2^{93}=9903520314283042199192993792\ \ \ $ but my computer confirmed that $2^n$ has a 6 for $n$ from $94$ up to at least $216927\ $. LaBird also provided a link to a related question a special case of which suggests that $2^n$ has all digits for $n\ge169$ (confirmed for $n$ up to $10000$)
Biggest powers NOT containing all digits.

My computer also tells me that the last $60$ digits of $2^{160021}$ have no $6$, but the last $100$ do.
The last $100$ digits of $2^{52613}$ have no $6$ (though $2^{52613}$ itself does).
The last $101$ digits of $2^{183137}$ have no $6$ (though $2^{183137}$ itself does).

@Soke says we know that the $m$-many last digits do cycle with period $4 \cdot 5^{m-1}$ (I didn't know and have yet to verify it for myself or find a reference, but I have no reason to doubt it). Does that make Q.1 appear easy (may boil down to a tedious verification)? On the other hand @Elaqqad suggests similarity to a hard problem of Erdős open since 1979, that the base-$3$ representation of $2^n$ contains a $2$ for all $n>8$. I found the following two related links:
http://mathworld.wolfram.com/Ternary.html and
https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/

@Elaqqad also provided a link to MSE question discussing the related conjecture that all powers of $2$ greater than $2^{86}$ have a zero in their base $10$ representation
Status of a conjecture about powers of 2

Answer 1

Q1: This is easy to state but to me seems hard to proof:

Each $i$-th place $d_i^{(n)}$ of $2^n=(d_{m-1}^{(n)}\cdots d_0^{(n)})_{10}$ develops with increasing $n$ similar to a pseudo random number generator, generating digits from $\{0,\ldots,9\}$. The length $m$ of this representation increases, introducing more generating places, so the probability that one of the $m$ places generates a $6$ is growing. Or the other way round: for large $n$ and thus large $m$ it becomes more and more unlikely that none of the $m$ places generates a digit $6$. All these places are coupled linear congruential generators $$ d_i^{(n+1)} = (2 d_i^{(n)}+c_i^{(n)})\mbox{ mod }10 $$ with different initial values and different couplings via the carry $c$ between places. Each can have a different initial run and a different cycle of generated digits with different cycle length. I am not sure if one can analyze them combinatorically to come up with a lower limit $k$ For $n$, or if this line of thought stays probabilistic.