20.2.1  For integers $m, n$ with $n>m \geqslant 2$, if the set $\{m, m+1, \cdots, n\}$ is arbitrarily divided into two

subsets, at least one subset contains $a, b, c$ (not necessarily distinct) such that $a^{b}=c$. Find the minimum value of $n$, $f(m)$.
Parse $A=\left\{m, m+1, \cdots, m^{m}-1\right\} \cup\left\{m^{m+1}, m^{m+1}+1, \cdots, m^{m^{m+2}}-1\right\}$, $B=\left\{m^{m}, m^{m}+1, \cdots, m^{m^{n+1}}-1\right\}$.

Since $\left(m^{m}\right)^{m^{m}}>m^{m^{m+1}}-1, m^{m}>m^{m}-1,\left(m^{m}-1\right)^{\left(m^{m}-1\right)}m^{m^{m+2}}-1$, it follows that $A \cup B$ does not meet the requirements, so $f(m) \geqslant m^{m+2}$.

If $f(m)<m^{m^{m+2}}$, assume without loss of generality that $m \in A$, then $m^{m} \in B,\left(m^{m}\right)^{m^{m}}=m^{m^{m+1}} \in A,\left(m^{m^{m+1}}\right)^{m}$ $=m^{m^{n+2}} \in B$, and $m^{\left(m^{n+1}\right)}=m^{m^{+1}},\left(m^{m}\right)^{m^{m+1}}=m^{m^{m-1}}$, so $m^{m+1} \notin A \cup B$, which is a contradiction! Therefore, $f(m)=m^{m+2}$.