Harvard-MIT Mathematics Tournament

Solution:

Suppose the number $n$ has $k+1$ digits, the first of which is $d$. Then the number is at least $d \cdot 10^{k}$. On the other hand, each of the digits after the first is at most $9$, so the product of the digits is at most $d \cdot 9^{k}$. Thus, if $n$ equals the product of its digits, then
$$d \cdot 10^{k} \leq n \leq d \cdot 9^{k}$$
which forces $k=0$, i.e., the number has only one digit. So $n=9$ is clearly the largest possible value.