\[a \downarrow b = a^b\] \[a \downarrow^n 1 = a\] \[a \downarrow^{n + 1} (b + 1) = (a \downarrow^{n +1} b) \downarrow^n a ext{ (otherwise)}\] \(a \downarrow^n b\) is a shorthand for \(a \downarrow\downarrow\cdots\downarrow\downarrow b\) with n down-arrows. When \(n = 2\), \(a \downarrow\downarrow b = a^{a^{b-1}}\). The inequality \(a \downarrow\downarrow a = a^{a^{a-1}} < a^{a^a} = a \uparrow\uparrow 3\) is useful when bounding down-arrows in terms of up-arrows. It can be shown that \(a \downarrow^{2n-1} b \ge a \uparrow^n b\) for \(a, b, n \ge 1\).