ISL 1989 Problem 79

Given two natural numbers $w$ and $n,$ the tower of $n$ $w's$ is the natural number $T_n(w)$ defined by

$$T_n(w) = w^{w^{\cdots^{w}}},$$

with $n$ $w's$ on the right side. More precisely, $T_1(w) = w$ and $T_{n+1}(w) = w^{T_n(w)}.$ For example, $T_3(2) = 2^{2^2} = 16,$ $T_4(2) = 2^{16} = 65536,$ and $T_2(3) = 3^3 = 27.$ Find the smallest tower of $3's$ that exceeds the tower of $1989$ $2's.$ In other words, find the smallest value of $n$ such that $T_n(3) > T_{1989}(2).$ Justify your answer.