TITLE: Proving that $f(n)=n$ if $f(n+1)>f(f(n))$
QUESTION [25 upvotes]: How can we prove that if $f:\mathbb{N}\rightarrow\mathbb{N}$ is a function so that $f(n+1)>f(f(n))$ for all $n\in\mathbb{N}$ then $f(n)=n$ for all $n\in\mathbb{N}$?
This is problem 6 from the IMO 1977. I found it in this book.

REPLY [3 votes]: Another proof :

We first show the property $P_n$ is true for all $n$
$\\$ $P_n : (\forall k \geq n \implies f(k) \geq n)$
Proof by Induction :
(i) $P_0$ is true, since $\forall k \geq 0 ( f(k) \geq 0)$
(ii) $\forall n$, if $P_n$ is true, then
$\forall k \geq n+1 ; k-1 \geq n$ then $f(k-1) \geq n$ then $f(f(k-1)) \geq n$ and
$f(k) > f(f(k-1)) \geq n$ thus $f(k) \geq n+1$
Therefore $P_{n+1}$ is true.

$\forall n$, since $P_n$ is true, take $k=n$, then $f(n) \geq n$

Then $\forall n, f(f(n)) \geq f(n)$ then the starting hypothesis yields:
$\\$ $f(n+1)>f(n)$ then $f$ is strictly growing.
Thus the starting hypothesis implies : $\forall n, n+1> f(n)$

Thus for all $n$, $n+1>f(n)>=n$. Hence, $f(n) = n$