Cubical Enigma

Note that for sufficiently large $a_n$ , the next number in the sequence is on the order of the number of digits in the current number (i.e: $a_{n+1} = O(log(a_n))$ ). Thus, for some K, $a_{n+1}<a_n$ for all $a_n>K$ . Thus, at some point $a_n \le K$ for all future $a_n$ . Because all $a_n$ are positive integers less between 1 and K, by the pigeonhole principle the sequence will repeat itself.