No. Denote $m$ the number of digits in $a_n$. We have that $a_{n+1}<\lceil _{10}\log(a_n)\rceil.9^{2005}$ (replacing all digits with 9), so if $a_n\ge10^{2005}$ then $a_{n+1}<a_n$. Now, $a_k\ge10^{2005}$, is $a_{k+1}\ge10^{2005}$? If yes, go on until you're below $10^{2005}$. (which will always occur as it's decreasing) If not, you also have another number below $10^{2005}$. So there are infinitely many numbers below $10^{2005}$, so they cannot all be distinct by pigeonhole.