The following question appeared in IMO 1977:
"Let $f(n)$ be a function defined on the set of all positive integers and with all its values in the same set. Prove that if $$f(n+1)>f[f(n)]$$ for each positive integer n then prove that $$f(n)=n$$ for each n"
I was able to prove that f has a unidue minimum at $n = 1$.
However, the solutions I find ask me to prove that $$f(1) < f(2) < f(3) <....$$The solution here :https://mks.mff.cuni.cz/kalva/imo/isoln/isoln776.html proceeds by induction.
I see this part here in the solution:
"Suppose $S_n$ is true. Take $m > n+1$. Then $m-1 > n$, so by $S_n$, $f(m-1) > f(n)$. But $S_n$ also tells us that $f(n) > f(n-1) > ... > f(1)$, so $f(n) ≥ n - 1 + f(1) ≥ n$. Hence $f(m-1) ≥ n+1$. So $f(m-1)$ belongs to ${ n+1, n+2, n+3, .. }$. But we are given that $f(m) > f(f(m-1))$, so $f(m)$ is not the smallest element of $ {{ f(n+1), f(n+2), f(n+3), ... }} $.
The next statement tells that since there does exist a minimum, the minimum is $f(n+1)$. How is that concluded?