Proving that $f(n)=n$ if $f(n+1)>f(f(n))$
Claim 1: If $f(k) = 0$ then $k =0$.
Proof: Suppose not. Then there exists $k$ such that $0 = f(k) > f(f(k-1))$, which is not possible, as $f: \mathbb{N} \mapsto \mathbb{N}$.
Claim 2: $f(0) = 0$.
Proof: Let $S = \{f(k) | k > 0\}$. Let $a$ be the smallest number in $S$. Then there exists $k >0$ such that $a = f(k) > f(f(k-1))$. But this means $f(k-1) = 0$. Thus $k=1$, and $f(0) = 0$.
Claim 3: $f(n) = n$.
Proof: Assume, for all $0 \leq m < n$, that $f(k) = m$ iff $k = m$. Now we proceed as in the proofs of Claims 1 and 2.
If $f(k) = n$, then $f(f(k-1)) < n$, which means $k-1 = f(k-1) = f(f(k-1)) < n$. So $k < n+1$, which means that $k = n$.
Let $S = \{f(k) | k > n\}$. Let $a$ be the smallest number in $S$. Thus there exists $k >n$ such that $a = f(k) > f(f(k-1))$. Therefore $f(k-1) \leq n$. But if $f(k-1) < n$, then $f(k-1) = k-1$, and so $k \leq n$. But $k>n$. Therefore, $f(k-1) = n$, and so $k= n+1$ and $f(n) = n$.
Another proof :
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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.
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$\forall n$, since $P_n$ is true, take $k=n$, then $f(n) \geq n$
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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, $f(n) = n$
Attempted solution (Improvements/corrections are greatly welcome!):
Clearly, the function isn't constant on any interval. So it must be either strictly increasing or strictly decreasing.
Suppose the function isn't injective. Then there exist $a,b \in \mathbb{N}$ such that $f(a) = f(b)$, but $a \neq b$. So either $a>b$ or $a<b$. WLOG, let $a>b$. Then $a = b+m$, for some $m \in \mathbb{N}$. Now, $f(a) = f(b+m) > f(f(b+(m-1)) > f(f(f(b+m-2) > ... >f(f(...(f(b))...)$. Now, suppose, WLOG that $f$ is increasing (if $f$ is decreasing, then there is also a contradiction). Then $f(f(...(f(b))...) > f(b)$. But $f(b) = f(a)$. Thus, we have a contradiction. So, $f$ must be injective.
Now, apply $f^{-1}$ on both sides, and you get that $n+1 > f(n)$, $\forall n \in \mathbb{N}$.
Clearly, if $ \forall n \in \mathbb{N}$, $f(n) = a*n+b$, where $a \neq 1$ and $b \neq 0$, then it would contradict $n+1 > f(n)$, $\forall n \in \mathbb{N}$.
Now it's just left to see what happens if there exists an $n'$ where $f(n') \neq n'$. In such a case, $f(n')>n'$ or $f(n')<n'$. Check that both cases still lead to contradictions.
I think if you can somehow rule out that $f$ is decreasing, then this might work.