Which number is larger,

$$
1997^{1998^{1999}} \quad \text { or } \quad 1999^{1998^{1997}} ?
$$
$$
A=1997^{1998^{1999}}=1997^{1998^{(1997+2)}}=1997^{1998^{2} \cdot 1998^{1997}}=\left(1997^{1998^{2}}\right)^{1998^{1997}}
$$

Let $a=1997^{1998^{2}}, B=1999^{1998^{1997}}, b=1999$.

Since $1997^{2}>1999$, it follows that $1997^{1998^{2}}>1999$, that is, $a>b$, and since both are numbers greater than 1, raising them to the same power preserves the inequality. Therefore, $A=a^{1998^{1997}}>B=b^{1998^{1997}}$.

Thus, the number on the left is the larger one.

Based on the solution by János Nagy 725 (Eger, Dobó I. Gymn., 9th grade) and József Sándor Karácsonyi (Szekszárd, Garay J. Gymn., 10th grade).

Remarks. 1. Several used logarithms.

2. Viktor Harangi (Fazekas M. Főv. Gyak. Gymn., 9th grade) similarly proved, for all integers $k \geq 3$, that $(k-1)^{k^{k+1}}>(k+1)^{k^{k-1}}$.