Probability that the first digit of $2^{n}$ is 1
Solution 1:
Hint 1: Show that there is always a power of 2 that has $k$ digits and starts with 1. (For $k=1$, I'm including $2^0=1$.) Use Shreevastsa's hint about logarithms.
Hint 2: Show that there is at most 1 power of 2 that has $k$ digits and starts with 1.
Hence, there are exactly $k$ powers of 2 from 1 to $2 \times 10^k$ that start with 1.
For any $n$, $2^n$ has $ \lfloor n \log 2 \rfloor + 1$ digits, hence $a_n = \lfloor n \log 2 \rfloor$, so $a_n < n \log 2$, which gives us the RHS.
For the LHS, show that $\lceil n \log 2 \rceil - 1 = a_n $.