October 12th
Today I learned that the leading digits of Fibonacci numbers $F_n$ follow Benford's Law. (Yesterday I learned $9^k$ follows Benford's Law.) The main lemma is that $\{\log F_n\}$ is equidistributed in $[0,1],$ where we are taking $\log$s base $10.$ (The base isn't important, so I prefer not to write it.) In particular, we use Binet's Formula to write\[\log F_n=\log\left(\frac{\varphi^n-(-1)^n\varphi^{-n}}{\sqrt5}\right)=n\log\varphi+\log\left(1-(-1)^n\varphi^{-2n}\right)-\log\sqrt5.\]Note $\log\sqrt5$ is a translation and will not affect the distribution$\pmod1,$ so we may ignore it. We're left with\[n\log\varphi+\log\left(1-(-1)^n\varphi^{-2n}\right).\]We already know that $\{n\log\varphi\}$ is equidistributed by the Equidistribution theorem, so we hope that the other term is sufficiently small; we show that it is. I guess we have to show that $\log\varphi$ is irrational, but $\log\varphi=\frac pq\in\QQ$ would imply that $\varphi^q=10^p.$ However, no $\varphi^\bullet$ is an integer by adding in their conjugate power, which is less than $1,$ so this is a contradiction.
Note that as $n\to\infty,$ we have $\log\left(1-(-1)^n\varphi^{-2n}\right)\to0,$ so we may fix a bound $N'$ so that $n \gt N'$ forces $\log\left(1-(-1)^n\varphi^{-2n}\right) \lt \varepsilon,$ for an arbitrary $\varepsilon.$ It follows that\begin{align*} & \left\{n \lt N:\{n\log\varphi+\log\left(1-(-1)^n\varphi^{-2n}\right)\}\in(x,y)\right\} \\ \subseteq & \left\{N' \lt n \lt N:\{n\log\varphi\}\in(x+\varepsilon,y-\varepsilon)\right\},\end{align*}and\begin{align*} & \left\{n \lt N:\{n\log\varphi+\log\left(1-(-1)^n\varphi^{-2n}\right)\}\in(x,y)\right\} \\ \supseteq & \left\{N' \lt n \lt N:\{n\log\varphi\}\in(x-\varepsilon,y+\varepsilon)\right\},\end{align*}where we are somewhat ignoring looping errors with $x$ and $y$ close to $1$—we could just send $\varepsilon$ sufficiently small so these don't occur. It follows that\[\lim_{N\to\infty}\frac{\#\left\{n \lt N:\{n\log\varphi+\log\left(1-(-1)^n\varphi^{-2n}\right)\}\in(x,y)\right\}}N\in[y-x-2\varepsilon,y-x+2\varepsilon]\]because we already know $\{n\log\varphi\}$ is equidistributed. Taking $\varepsilon\to0$ implies that the limit is $y-x,$ implying that $\log F_n$ is equidistributed as well.
To finish, we show that the equidistributed nature of $\log F_n$ implies that the leading digit of $F_n$ follows Beford's Law. Indeed, the leading digit of $F_n$ is $d\in[0,10)\cap\ZZ$ if and only if\[d\cdot10^\bullet\le F_n\le(d+1)\cdot10^\bullet\]for some suitable power of $10$ named $10^\bullet.$ Taking logs, this is equivalent to\[\bullet+\log d\le\log F_n\le\bullet+\log(d+1).\]However, we know that $\bullet$ is an integer, so taking fractional parts is legal, implying that this is equivalent to\[\log d\le\{\log F_n\}\le\log(d+1).\]In particular, we require $\log F_n\in[\log d,\log(d+1)],$ which occurs with probability $\log(d+1)-\log d$ because we just showed that $\log F_n$ is equidistributed. But this is Benford's Law, so we are done here.