Consider the sequence $1,2,4,8,...,a^n = 2^n,...$ of all the powers of $2$. Prove that, given any digit $i ∈ {1,...,9}$, there exist infinitely many values of $n$ for which $a^n$ starts with that digit.
It only comes to my mind to use Poincaré's Recurrence Theorem, but I can not do that. Can anyone give a tip
Brute force approach:
Write all terms in scientific notation, i.e. as $a\cdot 10^k$ for integer $k$ and $1\leq a<10$. Assume for contradiction that only finitely many terms of the sequence begin with $3$. In other words that only finitely many terms have $a\in[3,4)$ in our scientific notation.
That also means that only finitely many terms can have $a\in[6,8)$, as all such terms require the term before them to begin with $3$. Continuing with the same idea, we get that only finitely many terms have $a\in[1.2, 1.6)$. And so on.
It doesn't take too many of these intervals together to cover all of $[1,10)$, meaning there can only be finitely many terms, which is clearly a contradiction. The same argument holds for any other starting digit.