Answer: not. We can always find such $k$ that $x= 10^k-1$ and $x,x^2,...,x^n$ begin with $9$ Proof: Obvious, that enough to prove, that $x^n$ begin with $9$ so $(10^k-1)^n>9*10^{kn-1}$ $(10^k-1)^n=10^{kn}-n10^{k(n-1)}+...>10^{kn}-n10^{k(n-1)}-C_{[\frac{n}{2}]}^{n}(1+10^k+...+1+10^{k(n-2)})>10^{kn}-n10^{k(n-1)}-C_{[\frac{n}{2}]}^{n} 10^{k(n-1)}$ Set $k> \log_{10}{(C_{[\frac{n}{2}]}^{n}+n)}+1$ Then $10^{k-1}>C_{[\frac{n}{2}]}^{n}+n$ and $10^{kn}-n10^{k(n-1)}-C_{[\frac{n}{2}]}^{n} 10^{k(n-1)}>10^{kn}-10^{k-1}10^{k(n-1)}=9*10^{kn-1}$