find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$ find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$
My approach was
$$x^{3(x^x-1)} \lt x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$$
and if x is 5 -> $5^{3(5^5-1)}=5^{9372} \gt 1990^{1990}$
so $x\in \{1,2,3,4\}$,and I tried to check in cases when x=1, x=2,x=3,x=4 but I couldn't proceed.

First note the prime factorization of $1990 = 2 \times 5 \times 199$.
Hence at least one of $x,y,z$ must be divisible by $199$, from the equation $x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$.
Ignore the order for now and suppose $199 \mid x$. Then $x^{x^y-1} \ge 199^{199^y-1}$ and $z^{z^x-1} \ge z^{z^{199}-1}$.
For $y \ge 2$, $199^{199^y-1} \ge 199^{199^2-1} \gg 1990^{1990}$ (by observing logarithms.) Hence $y=1$.
Similarly we see that for $z \ge 2$, $z^{z^{199}-1} \ge 2^{2^{199}-1} \gg 1990^{1990}$. Hence $z = 1$ as well.
This reduces the equation to
$$x^{x-1} = 1990^{1990}$$
which has no solutions, since $2,5,199 \mid x$.
