We know that $2+2 = 2 \times 2 = 2^2$. Is this true for all successive hyperoperations?

Question

Given that multiplication is repeated addition and exponentiation is repeated multiplication, if we were to continue onwards, what would the result be for successive hyperoperations? If we take the next step after exponentiation which is tetration, it happens to be that $2 \uparrow 2$ does indeed give us $4$ as the result. With a pattern established, my initial hypothesis is that the result of any hyperoperation applied between $2$ and $2$ is going to be equal.

What would be the correct way to go about proving whether this is true or not in a rigorous manner? I'm guessing I probably need to define some kind of higher order function for the sequence of hyperoperations, and then find some relationship between an operation and the operation after it in order to determine my result?

This isn't a homework problem by the way, though I imagine it could be one. It's just something that came to mind, that I thought would be interesting to try and find an answer to.

Answer 1

From the definition of Knuth arrow, you can see easily that: $$ 2\uparrow^{n}2=2\uparrow^{n-1}2 $$ You just apply the $\uparrow^{n-1}$ to two twos. And you can do this over and over until you get $2\uparrow2$ which is 4. So it is true.

Answer 2

Definition. Define $\uparrow:\mathbb N\times\mathbb N\times\mathbb N\to\mathbb N$ recursively.

$m\uparrow^0 n:=mn\\ m\uparrow^{k+1} 0:=1\\ m\uparrow^{k+1} n+1:=m\uparrow^{k}(m\uparrow^{k+1}n)$

Lemma 1. $\forall m\forall k.m\uparrow^k 1=m$

Proof. If $k=0$, $m\uparrow^0 1:=m1=m$. Assume $m\uparrow^k1=m$. Then $m\uparrow^{k+1}1=m\uparrow^k(m\uparrow^{k+1}0)$. By Definition of $\uparrow$ we conclude $m\uparrow^{k+1}1=m\uparrow^k 1$.

$\square$

Theorem. $\forall k.2\uparrow^k 2=4$

Proof. If $k=0$, $2\uparrow^0 2:=2\cdot 2=4$. Assume $2\uparrow^k2=4$. Then $2\uparrow^{k+1}2=2\uparrow^k(2\uparrow^{k+1}1)$. By Lemma 1 of $\uparrow$ we conclude $2\uparrow^{k+1}2=2\uparrow^k 2$.

$\square$