TITLE: Unconventional types of induction
QUESTION [18 upvotes]: Induction is one of the most common tools is mathematics, and everybody knows the ordinary induction and the strong induction. However, in some proofs induction is applied in an unexpected and elegant way. This can happen in two ways:

The proof uses a special form of induction.  
The variable that is inducted on is surprising. 

To clarify what 1 and 2 mean, let me give an example of both.

Cauchy's proof of the arithmetic-geometric inequality $\frac{x_1+\cdots+x_n}{n}\geq \sqrt[n]{x_1\cdots x_n}$ proceeds by showing that the case $n=2^k$ implies the case $n=2^{k+1}$ and that the case of $n=k$ implies the case $n=k-1$. This is an unconventional type of induction that turns out to be suitable for this theorem (though it is not the only way to prove it).
A proof of the van der Waerden theorem (given here) goes as follows: Assume that $W(r,k-1)$ exists. By induction on $n$ we see that there exists a number $N=N(r,k,n)$ such that if the set $[1,N]\cap \mathbb{N}$ is colored with $r$ colors, one can either find a monochromatic arithmetic progression of length $k$ or $n$ arithmetic progressions of length $k-1$ each of which is monochromatic but has a different color. Then taking $n=r+1$ we get that $W(r,k)$ is finite and the ordinary induction on $k$ continues.

Question: What other examples are there of proofs of famous and non-trivial results where induction is crucial in the argument and it is of the form 1 or 2? 
The types of induction include for example induction on prime numbers, induction on the rational numbers, inductions based on the parity of the variable, inductions where the cases are not proved in an incresing order (as in example 1 above), and so on. 
By a surprising variable of induction I mean one that is not given in the theorem and adding this to the theorem is not obvious (so a non-example would be proving an inequality of three variables by inducting on the number of varibles).

REPLY [2 votes]: Laver tables often involve unconventional types of induction. In fact, constructing the Laver tables and proving that they are self-distributive involves an unusual double and triple induction.
Consider the following Lemmas.

$\textbf{Lemma 1}$ For all $N$ there exists a unique binary operation
  $*$ on $\{1,...,N\}$ such that 

$x*1=x+1$ whenever $x<N$
$N*x=x$ for all $x$ and
$x*y=(x*(y-1))*(x+1)$ whenever $x<N$ and $y>1$.


One proves Lemma 1 for each $x*y$ by a double induction which is descending on $x$ and for each $x$ the induction is ascending on $y$.
In fact, many results about Laver tables are proven using a descending then ascending double induction.

$\textbf{Lemma 2}$ For all $N$ the algebra $(\{1,...,N\},*)$ satisfies
  the self-distributivity identity $x*(y*z)=(x*y)*(x*z)$ if and only if
  $x*N=N$ whenever $x\in\{1,...,N\}.$

The direction $\rightarrow$ in Lemma 2 is straightforward, but the direction $\leftarrow$ is by descending induction on $x$, then for each $x$ one proceeds by ascending induction on $y$ and for each $y$, one uses ascending induction on $z$.
Notice how one uses a descending induction for each variable in $x*(y*z)$ to the left of the operation $*$ and one uses an ascending induction for each variable to the right of the operation $*$. Similarly, one uses a descending induction for $x$ in $x*y$ in Lemma 1 and one uses an ascending induction on $y$ in $x*y$ in Lemma 1 because $x$ is on the left side of $*$ and $y$ is on the right side of $*$.
Lemma 2 is needed to prove the following theorem.

$\textbf{Theorem}$ The algebra $(\{1,...,N\},*)$ satisfies the
  self-distributivity identity $x*(y*z)=(x*y)*(x*z)$ if and only if
  $N=2^{n}$ for some $n$.

The self-distributive algebra $(\{1,...,2^{n}\},*)$ is known as the $n$-th classical Laver table.