DFA for accepting all binary strings of form power of $n$ (not divisible by $n$) i.e. $n^k$ for given $n$ We can form DFA accepting binary numbers divisible by $n$. 

For example DFA accepting binary numbers divisible by 2 can be formed as follows:



Similarly DFA accepting binary numbers divisible by 3 can be formed as follows:


We can follow a well defined procedure to form these types of DFAs.
However can there be any well defined procedure or better to say logic for forming DFAs accepting numbers of of the form $n^k$?

For example, let us consider DFA accepting all numbers of the form $2^k$.
This language will be $\{1,10,100,1000,...\}$, thus have regex $10^*$. We can form DFA as follows:


I tried forming DFA for $3^k$ and similar ones? But was not able to do so. Or is it just that its pattern of $2^n$ binary equivalents which was making it possible to create DFA and we cannot form DFA accepting all binary numbers of the form $n^k$ for specific $n$?
Here is a quick and dirty proof using Pumping Lemma that language $L$ consisting of $3^n$ in binary is not regular (note: it is regular if represented in ternary, so representation is important).

I will use the notation from Wikipedia article re Pumping Lemma. Assume for contradiction that $L$ is regular. Let $w \in L$ be any string with $|w| \ge p$ (pumping length). By Pumping Lemma, write $w = x y z$ with $|y| \ge 1, |xy| \le p$ and for all $i \ge 0$ $xy^i z \in L$. I will write $x$, $y$, and $z$ also for numerical values of corresponding parts, and $|x|, |y|, |z|$ for their lengths in $w$. Numerically we have $w = 3^{k_0}$ for some $k_0 \in \mathbb{N}$. At the same time we have numerically $w = z + 2^{|z|} y + 2^{|z|+|y|}x$. Thus, we have

$$ z + 2^{|z|}y+2^{|z|+|y|} x = 3^{k_0}$$

Now, let's pump $w$ to get for all $i \ge 0$

$$ z + 2^{|z|} y \left( \sum_{j=0}^{i-1} (2^{|y|})^j \right) + 2^{|z|+i|y|} x = 3^{k_i}, $$

where $k_0 &lt; k_1 &lt; k_2 &lt; \ldots$. Simplifying we get for $i \ge 1$

$$ z + 2^{|z|} y (2^{i |y|} - 1) / (2^{|y|}-1) + 2^{|z|+i|y|} x = 3^{k_i}.$$

Let $C = z - 2^{|z|} y / (2^{|y|}-1)$. Then we have

$$ 3^{k_i} = 2^{|z|+i |y|} y / (2^{|y|}-1) + 2^{|z|+i|y|} x + C. $$

Now, observe that

$$ 3^{k_i} - 3^{k_{i-1}} = (2^{|y|}-1)(3^{k_{i-1}} - C). $$

Therefore, we have $C (2^{|y|}-1) = 3^{k_{i-1}} (2^{|y|} - 3^{k_i - k_{i-1}}).$ Note that $|2^{|y|} - 3^{k_i - k_{i-1}}| \ge 1$. Thus, on one hand, the absolute value of the right hand side grows at least as $3^{k_{i-1}}$, which goes to infinity with $i$. On the other hand $C(2^{|y|}-1)$ is independent of $i$ and is a constant. This gives a contradiction.