The great Catholic mathematician Edward Nelson does not believe in completed infinity, and does not believe that arithmetic is likely to be consistent. These beliefs are partly motivated by his faith: he says arithmetic is a human invention, and compares believing (too strongly) in its consistency to idolatry. He also has many sound mathematical insights in this direction -- I'll summarize one of them here.
http://www.mediafire.com/file/z3detbt6int7a56/warn.pdf
Nelson's arguments flow from the idea that, contra Kronecker, numbers are man-made. He therefore does not expect inconsistencies to have consequences that play out in natural or divine processes. For instance, he does not expect you to be able to count the dollars in a stack of 100 dollars and arrive at 99 dollars. But it's been known for a long time that if one can prove any contradiction, then one can also prove that a stack of 100 dollars has no more than 99 dollars in it. The way he resolves this is interesting.
The Peano axioms for the natural numbers are these:
1. Zero is a number
2. The successor of any number is a number
3. Zero is not the successor of any number
4. Two different numbers have two different successors
5. If a given property holds for zero, and if it holds for the succesor of x whenever it holds for x, then it holds for all numbers.
Nelson rejects the fifth axiom, induction. It's the most complicated of the axioms, but it has another thing going against it: it is the only one that seems like a claim that could be either true or false. The first four axioms read like someone explaining the rules of a game, like how the pieces in chess move. Induction is more similar to the fact that the bishop in chess can only move on half the squares -- this is a theorem about chess, not one of the rules. Nelson believes that the fifth axiom needs to be, and cannot be, supported.
A common way to support induction is via the monologue: "It's true for zero. Since it's true for zero it's true for one. Since it's true for one it's true for two. Continuing like this we can show that it's true for one hundred and for one hundred thousand and for every natural number." It's hard to imagine actually going through this proof for very large numbers -- this is Nelson's objection.
What is arithmetic like if we reject induction? First, we may make a distinction between numbers we can actually count to (call them counting numbers) and numbers that we can't. Formally we define counting numbers as follows: 0 is a counting number, and if x is a counting number then so is its successor. We could use the induction axiom to establish that every number is a counting number, but without it we cannot.
A small example of a number so large we might not be able to count that high is the sum of two counting numbers. In fact without induction we cannot establish that x+y is a counting number from the facts that x and y are counting numbers. So we cut out a smaller class of numbers called additionable numbers: x is additionable if x + y is a counting number whenever y is a counting number. We can prove theorems about additionable numbers: for instance every additionable number is a counting number, the successor of an additionable number is additionable, and in fact the sum of two additionable numbers is an additionable number.
If we grant the induction axiom, these theorems lose their interest: every number is a counting number and an additionable number. Paraphrasing Nelson: the significance of these theorems is that addition is unproblematic even if we are skeptical of induction.
We can go further. It cannot be proved that the product of two additionable numbers is additionable. We therefore introduce the smaller class of multiplicable numbers. If whenever y is an additionable number x.y is also additionable, then we say that x is a multiplicable number. It can be proved that the sum and product of any two multiplicable numbers is multiplicable. Nelson closes the article I linked to:
The proof of the last theorem [that the product of two multiplicable numbers is multiplicable] uses the associativity of multiplication. The significance of all this is that addition and multiplication are unproblematic. We have defined a new notion, that of a multiplicable number, that is stronger than the notion of counting number, and proved that multiplicable numbers not only have successors that are multiplicable numbers, and hence counting numbers, but that the same is true for sums and products of multiplicable numbers. For any specific numeral SSS. . . 0 we can quickly prove that it is a multiplicable number.
But now we come to a halt. If we attempt to define “exponentiable number” in the same spirit, we are unable to prove that if x and y are exponentiable numbers then so is x ↑y. There is a radical difference between addition and multiplication on the one hand and exponentiation, superexponentiation [what is commonly denoted ^^ here], and so forth, on the other hand. The obstacle is that exponentiation is not associative; for example, (2↑2)↑3 = 4↑3 = 64 whereas 2↑(2↑3) = 2↑8 = 256. For any specific numeral SSS...0 we can indeed prove that it is an exponentiable number, but we cannot prove that the world of exponentiable numbers is closed under exponentiation. And superexponentiation leads us entirely away from the world of counting numbers.
The belief that exponentiation, superexponentiation, and so forth, applied to numerals yield numerals is just that -- a belief.
I've omitted his final sentence.
Actually, for moderately large numbers it's easy:
"Since it's true for numbers one higher than numbers it's true for, it's true for numbers two higher than numbers it's true for." "....two....four...." "....four....eight..." "....eight....sixteen..."
Of course, if that's not fast enough you can try:
"Since it's true for gaps double the size of gaps it's true for, it's true for gaps quadruple the size it's true for" "...quadruple... sixteen times..." "...sixteen times...256 times..."
So, we've gone from N repetitions proving it for numbers up to N, to N repititions proving it for numbers up to 2^N, to N repititions proving it for numbers up to 2^(2^N). And we could continue.
Very nice. It's interesting to think about how hard it would be to prove the proposition "3^^^3 is a counting number" using your trick but not using induction.