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Even more is true: the leading digits of $p^n$ can be any sequence you want. Note how $$\text{the leading digits of }p^n\text{ are }\overline m$$ is the same as saying $$\overline m\cdot 10^k\leq p^n<(\overline m+1)\cdot 10^k\text{ for some }k\geq0.$$ Equivalently, $$\log_{10}(\overline m)\leq n\log_{10}(p)-k<\log_{10}(\overline m+1).$$ Indeed there ...
You get this because the golden ratio is a quadratic algebraic integer. Let $\tau=(1+\sqrt5)/2$ and $\overline{\tau}=(1-\sqrt5)/2$. These are the zeros of the polynomial $$p(x)=(x-\tau)(x-\overline{\tau})=x^2-x-1.$$ Let then $a/b$ be a rational number (in reduced form). Because $\tau$ is irrational we get that $p(a/b)\neq0$. But  p(a/b)=(\frac ...