Writing down a couple of powers of two, I saw that
$$2^{23} = 8388608$$
which seemed to contain an abnormally high amount of 8's. I then thought what if by choosing appropriate values of $k$ and $n$, I can get $D(k,n)$ as close to $1$ as I liked, where $D(k,n)$ is the percentage of digits in $2^n$ that are equal to $k$. For example,
$$D(8,23) = \frac{4}{7}$$
because there are $4$ instances of the digit $8$ in the number $2^{23}$ in base 10 (which has 7 digits). Of course, we have some trivial solutions like
$$D(4,2) = 1$$
because 2 squared is 4, but I'm interested in $n > 3$. After some computer simulation, however, it seems like this is probably definitely not the case, though I have no idea how to prove it either way.
(A similar conjecture that might be true is if by choosing an appropriate value of $n$, I can get $E(k,n)$ as close to $0.1$ as I like, where $E(k,n)$ is defined by:
$$E(k,n) = \text{minimum } D(k,n) \text{ achievable by varying k and keeping n constant}$$
I tried to approach this by looking at the 'period' of each digit. For instance, the last digit cycles with period $4$, (it goes 2, 4, 8, 6 then back to 2), and the second last digit cycles with period $20$, then period $100$, then period $500$, though a) I don't know how to explain the pattern in this cycling (has it got anything to do with the fact we are using base 10 and 10 = 2 times 5?) and b) I've got no idea how to use this fact to prove either conjecture
