Why is tetration considered the next step after exponentiation?

Question

Tetration is often stated to be the next step after exponentiation (see for example Wikipedia): $$\large a^{a^{a^{...^a}}}$$

Where the exponents are taken $b$ times from the top. I refer to the Wikipedia article and many questions on this site for clarification and references.

However, it seems to be incorrect, if we 'honestly' continue the sequence of operation. Let's do that.

For now we consider only $a,b \in \mathbb{N}$.

$$a \cdot b=a + \ldots + a \tag{b times}$$

$$a^b=a \cdot \ldots \cdot a \tag{b times}$$

And here is how I think we should continue:

$$\large a^{a^b}=\left(\left(\left(a\right)^a\right)^a \dots \right)^a \tag{b times}$$

This is the only 'honest' form of the next hyperoperation, in my opinion. We take the operation from the previous step, replace $b$ by $a$, then repeat the operation $b$ times. In no way tetration appears on this step (or not yet)!

Then for the next step we should go the same way, replace $b$ by $a$ (we get $a^{a^a}$), and repeat $b$ times:

$$\large a^{a^{ab}}=\left(\left(\left(a\right)^{a^a}\right)^{a^a} \dots \right)^{a^a} \tag{b times}$$

And here our 'power tower' stops growing for a while! Becasue of the usual rules of exponentiation, we have on the next step:

$$\large a^{a^{a^2b}}=\left(\left(\left(a\right)^{a^{a^2}}\right)^{a^{a^2}} \dots \right)^{a^{a^2}} \tag{b times}$$

Thus, to obtain even $$\large a^{a^{a^a}}$$ we have to make many more steps.

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I think the reason people call tetration the 'next' operation after exponentiation is the pure visual appeal of it, when expressed in the usual notation. It 'kind of' looks like repeated exponentiation, but it's really not.

Why is tetration considered the next step after exponentiation? Is there some flaw in my logic?

Answer 1

It comes down to left or right bias and loss of symmetry.

Rewrite your binary operation as $f(x, y)$ instead. The 3 times repetition to get the 'next' operation on $a$ with 3 could be $f(f(a, a), a)$ or it could be $f(a, f(a, a))$. This doesn't make a difference at first, but it does once you try to move up from powers. One way gives tetration and the other gives your result.

Answer 2

If we define the function $$f(n)=a^n$$ then tetration arises very natural. We have $$a\uparrow\uparrow b=f^b(1)$$

So, we start with $1$ and apply $f$ $b$ times. This way, we get $a$ after the first step , $a^a$ after the second step, $a^{(a^a)}$ after the third step and so on.

Finally, we arrive at the power tower containing $b$ $a's$.

Moreover, the usual convention to handle nested exponentiation is that $a^{b^c}$ is considered to be $a^{(b^c)}$ rather than $(a^b)^c$. This is also the way to get extremely large numbers (but this is , I admit, not a very convincing argument).