Show that there are infinitely many positive integers n so that 2^n starts with d Given the studies of rotational transformations and measurable transformations, let d be any positive integer. Show that there are infinitely many positive integers n so that $2^n$ starts with d.

The claim follows from the following Lemma in dynamics:
Let $\alpha$ be an element in $S^1$ of infinite order (i.e $\alpha^n\not=1$ for every $n$). Then the set $\{\alpha^n : n\in\mathbb{N}\}$ is dense in $S^1$.
How can we use the lemma? the number $2^n$ starts with $d$ if and only if there exists $k$ so that $d\cdot 10^k\leq 2^n<(d+1)\cdot 10^k$, as log is monotone this is true if and only if $log(d)+klog(10)\leq n log(2)< log(d+1) +klog(10)$, divide by $log(10)$ we have

The number $2^n$ starts with $d$ iff there exists $k$ so that $\frac{log(d)}{log(10)}+k\leq n \frac{log(2)}{log(10)}<\frac{log(d+1)}{log(10)}+k$.

This is true, if and only if $e^{2\pi i n \frac{log(2)}{log(10)}}$ lies in the interval between $e^{2\pi i \frac{log(d)}{log(10)}}$ and $e^{2\pi i \frac{log(d+1)}{log(10)}}$.
Use the Lemma above and the proof is complete.
