If $\beta=0.{a_1}^{k}{a_2}^{k}{a_3}^{k}\cdots\in\mathbb Q$, then $\alpha=0.a_1a_2a_3\cdots\in\mathbb Q$?

The general problem is this: We are given a (finite) alphabet $\Sigma$ and a map $f\colon \Sigma\to\Sigma ^*$. Via $a_1a_2\cdots \mapsto f(a_1)f(a_2)\cdots $ we can extend $f$ to the set $\Sigma^*$ of words, but even to the set $\Sigma^\omega$ of infinite strings over the alphabet $\Sigma$.

For $v,w\in\Sigma^*$ we introduce the relation is-suffix-of: $$ v\preceq w\iff \exists u\in \Sigma^*\colon uv=w$$

The question is:

Q: Is it possible for $\alpha\in\Sigma^\omega$ that $f(\alpha)$ is eventually periodic, but $\alpha$ is not?

The answer depends on $f$, of course. Let us assume that $f$ has the following property:

Property X. If $a,b\in\Sigma$, $w\in\Sigma^*$, $f(a)\preceq f(b)\preceq f(wa)$, then $a=b$.

Now assume $\alpha\in\Sigma^\omega$ is not eventually periodic and $f(\alpha)$ is eventually periodic. Say, for $p>0$ we have that $f(\alpha)_n=f(\alpha)_{n+p}$ for all $n>N$. For each $n\in\Bbb N$ with $|f(\alpha_1\cdots\alpha_{n-1})|>N$, define $\phi(n)=\langle \alpha_n,|f(\alpha_1\cdots\alpha_{n})|\bmod p\rangle$. Since $\phi$ takes only finitely many values, some value $\langle a_0,r_0\rangle$ occurs infinitely often. Then by property X, for all $n$ with $\phi(n)=\langle a_0,r_0\rangle$, $\phi(n-1)$ is either undefined or always takes the same value $\langle a_1,r_1\rangle$. By repeating the argument, for all such $n$, $\phi(n-2)$ is undefined or takes the same value $\langle a_2,r_2\rangle$, and so on. Again by finiteness, we will at some point find $\langle a_i,r_i\rangle=\langle a_j,r_j\rangle$ with $i< j$. From then on, we will have $\langle a_{i+k},r_{i+k}\rangle=\langle a_{j+k},r_{j+k}\rangle$ for all $k\ge 0$ and hence $\alpha$ is eventually periodic.

Thus we have shown:

If $f$ has property X, then the answer to question Q is "no".

Now we need to check if property X holds in the specific case where $\Sigma=\{0,1,2,3,4,5,6,7,8,9\}$ and $f(a)$ is the decimal representation of the number $a^k$. For odd $k$ this is clear because $\phi(10)=4$ and hence $a^k\equiv b^k\pmod{10}$ implies $a\equiv b\pmod{10}$, i.e., already $f(a)\preceq f(b)$ implies $a=b$.

For $k=2$, the $f(a)$ are $$0,1,4,9,16,25,36,49,64,81 $$ and we see that $f(3)$ is a suffix of $f(7)$ and also $f(7)$ is a suffix of $f(23)$. Therefore property X does not hold here (and indeed the answer to question Q is "yes", as is shown by letting $\alpha$ be a concatenation of $23$ and $7$ in a non-repetitive pattern.

For $k=4$, the $f(a)$ are $$0,1,16,81,256,625,1296,2401,4096,6561$$ Only $f(1)$ is suffix of other $f(a)$, namely of $f(3)$, $f(7)$, $f(9)$. But (by inspection) neither $8$ nor $240$ nor $656$ is obtainable as a suffix of some $f(w)$. Hence property X holds.

For $k=6$, the $f(a)$ are $$ 0, 1, 64, 729, 4096, 15625, 46656, 117649, 262144, 531441.$$ Again, only $f(1)$ is a suffix (and this will be the cse for all even exponents), but only of $f(9)$ and $53144$ is no obtainable as suffix of some $f(w)$. Hence property X holds.

For $k=8$, we have $$0,1, 256, 6561, 65536, 390625, 1679616, 5764801, 16777216, 43046721$$ and need only verify that $656$, $576480$, $4304672$ are not obtainable as suffix. Check.

The main conjecture is

Conjecture 1. Property X holds for all $k\ge 3$.

Also, we have a subordinate conjecture

Conjecture 2. If $k\ge 3$ and $f(a)$ is a suffix of $f(b)$ for two digits $a,b$, then $a=b$ or $a=1$.

Some work on this: Let $0\le a<b\le 9$. Then $d_n:=b^n-a^n$ follows the recursion $d_0=0$, $d_1=b-a$, $d_{n+1}=(a+b)d_n-abd_{n-1}$. We consider only the last $m$ digits of this sequence, i.e., work modulo $10^m$ for various $m$. As there are only finitely many values, the sequence is eventually periodic. We can simplify mattes by working modulo $5^m$: We are only interested in cases where $a,b\notin\{0,5\}$. Hence $ab$ is invertible $\pmod {5^m}$ and the sequence is immediately periodic and more specifically $d_n\equiv 0$ implies $d_{rn}\equiv 0$ for all $r\in\Bbb N_0$. In other words, there exists $p\in \Bbb N_0$ such that $d_n\equiv 0\pmod{5^m}$ iff $p\mid n$. Write $p(a,b,m)$ for this $p$.

For given $2\le a<b\le 9$, start with $m_0=2$ and recursively let $$m_{n+1}=\lfloor \log_{10}a^{p(a,b,m_n)}\rfloor+1=\lfloor p(a,b,m_n)\log_{10}a\rfloor +1$$ be then number of digits of the smallest power of $a$ that agrees with the same power of $b$ in the last $m_n$ digits. Then conjecture 2 holds at least for digits $a,b$ for all $k$ with $4\le k<p(a,b,m_n)$. For all interesting pairs $(a,b)$, i.e., both of same parity and $5\notin\{a,b\}$, one verifies numerically that $m_1\ge 4$, $m_2>75$. (For a specific example: with $a=2, b=8$, we have $m_0=2$, $m_1=4$, $m_2=76$, $m_3=15936395277450702718847601312677127001857150683576647$). Also, one verifies $p(a,b,75)\ge4235164736271501695341612503398209810256958007812500$ for all these $(a,b)$. This shows that conjecture 2 holds for $k\le 4235164736271501695341612503398209810256958007812500$.

In order to show conjecture 1 for the same range of $k$, it suffices to investigate if one can build $\frac{b^k-1}{10}$ for those $b>1$ with $b^k\equiv 1\pmod{10}$.