> [!NOTE] > For all $m\in \mathbb{Z}_{\geq 1}$ and $n\in \mathbb{Z}_{\geq 2},$ define $f_{1}(n)=n$ and $f_{m+1}(n)=n^{f_{m}(n)}.$ Then $f_{m}(n)<n\underbrace{!!!\dots!}_{\text{$m$ times}}<f_{m+1}(n).$ **Note** $f$ represents the [[Tetration|tetration]] operation. ###### Proof   Define $g_{0}(n)=n$ and for all $i \in \mathbb{N},$ $g_{i+1}(n)=(g_{i}(n))!$. WTS $f_{m}(n)<g_{m}(n)<f_{m+1}(n). \tag{1}$ When $m=1,$ this true: that is, for all $n\geq 2,$ $n<n!<n^n.$ Suppose the the left inequality in $(1)$ is true for $m=k$: that is, $g_{k}(n)>f_{k}(n).$ Then $g_{k+1}(n)= (g_{k}(n))! > n^{g_{k}(n)}>n^{f_{k}(n)}=f_{k+1}(n)$observing that if $t>2n^2,$ then using [[Exponential Lower Bound for Factorial]] ($t! > (\frac{t}{2})^{t/2}$), $t! >(n^2)^{t-n^2}=n^t n^{t-2n^2}>n^t$ and that for $m\geq 2, n\geq 3,$ $g_{m}(n)>2n^2$ thus, $(g_{m}(n))!>n^{g_{m}(n)}.$ For the RHS inequality in $(1)$, STS $g_{0}(n)g_{1}(n)\cdots g_{m}(n)<f_{m+1}(n) \tag{2}$holds true for $m$ and $n$ as above. First, we establish, by induction on $m,$ that $g_{0}(n)g_{1}(n)\cdots g_{m}(n)< n^{g_{0}(n)g_{1}(n)\cdots g_{m-1}(n)}\tag{3}$for all $m\geq{1}$ and $n\geq 3.$ When $m=1,$ $(3)$ reads $n \cdot n! < n^n$ which is clearly true. Now suppose $(3)$ holds for $m=k.$ Then $\begin{gathered} g_0(n)g_1(n)\cdots g_{k+1}(n) =g_0(n)g_0(n!)\cdots g_k(n!)<g_0(n)(n!)^{g_0(n!)g_1(n!)\cdots g_{k-1}(n!)} \\ <n(n!)^{g_1(n)\cdots g_k(n)}<(n\cdot n!)^{g_1(n)\cdots g_k(n)} \\ <(n^n)^{g_1(n)\cdots g_k(n)}=n^{g_0(n)g_1(n)\cdots g_k(n)}, \end{gathered}$and so by [[Induction Principle|mathematical induction]], $(3)$ indeed holds for all $m\geq{1}, n \geq 3.$ When $m=1,$ $(2)$ reads $n \cdot n! < n^n$ which is clearly true. Now suppose $(2)$ is true for $m=k.$ Then $g_0(n)g_1(n)\cdots g_{k+1}(n)< n^{g_{0}(n)g_{1}(n)\cdots g_{k}(n)}< n^{f_{m+1}(n)}= f_{m+2}(n)$which is exactly $(2)$ for $m=k+1.$ Thus, by mathematical induction, $(2)$ holds for $m$ and $n$ as above. **Note**: observe that It was simpler to prove a strengthened version of the RHS inequality.