User blog comment:AJZajac/Tetragorean Tuples/@comment-11227630-20171210084621

For all perfect square number m = n^2, $$m^m=(n^2)^{n^2}=n^{2n^2}=(n^n)^{2n}=\underbrace{n^n\times n^n\times\cdots n^n\times n^n}_{2n}$$ always hold.