nmmnthrowaway·3 lata temu·discussI think a more interesting problem to consider after solving n^m = m^n is solving n^m = m^n + 1 in the natural numbers.
nmmnthrowaway·3 lata temu·discussn^n * n^(m - n) = n ^ m = m ^ n, so n^n divides m^n, so n divides m.