The first step is to show that f(1) < f(2) < f(3) < ... . We do this by induction on n. We take $ S_n$ to be the statement that f(n) is the unique smallest element of { f(n), f(n+1), f(n+2), ... }. For m > 1, f(m) > f(s) where s = f(m-1), so f(m) is not the smallest member of the set {f(1), f(2), f(3), ... }. But the set is bounded below by zero, so it must have a smallest member. Hence the unique smallest member is f(1). So $ S_1$ is true. 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), ... }. But there must be a smallest element, so f(n+1) must be the unique smallest member, which establishes $ S_{n + 1}$. So, $ S_n$ is true for all n. So $ n \leq m$ implies $ f(n) \leq f(m)$. Suppose for some m, $ f(m) \geq m+1$, then $ f(f(m)) \geq f(m+1)$. Contradiction. Hence $ f(m) \leq m$ for all m. But since $ f(1) \geq1$ and f(m) > f(m-1) > ... > f(1), we also have f(m) ≥ m. Hence f(m) = m for all m

let $x_1$ such as that $f(x_1)$ be the minimum of $f(N) \implies f(x_1) > f(f(x_1-1))$ which is a contradiction if $x_1 \ne 1$ so $x_1=1$ and the function is injective for $f(1)$ which means if $f(a)=f(1) \iff a=1$. now take the 2nd smallest value of $f(N)$ and assume it occurs for $x_2$ then $x_2>1\implies f(x_2)>f(f(x_2-1))$ which implies that $f(f(x_2-1))=f(1)$ since $f$ is injective for $f(1)$ then $f(x_2-1)=1 \le f(1)$ and we also know that $f(1)\le f(x_2-1)$ then $f(1)=1$ and $x_2=2$ and now we shall utilize induction in order to prove the statement, statement: $f$ is injective for $A=\{f(1),f(2),...,f(n)\}$ and it reaches its $i^{th}$ smallest value with $i$ and $\forall i\in A -\{f(n)\}:f(i)=i$ we have already proved the it for $n=2$, now for the induction step, let $x_{n+1}$ be such as that $f(x_{n+1})$ be that $(n+1)$th smallest values in $f(N)$. so $f(x_{n+1}) > f(f(x_{n+1}-1)) \implies \exists k, 1\le k \le n: f(f(x_{n+1}-1))=f(k)$ $\implies x_{n+1}=k+1$ and if $k \le n-1$ we can see the contradiction with induction hypothesis so $k=n \implies x_{n+1}=n+1$ $f(x_{n+1}-1) =f(n) \ge n$ and $f(n+1) > f(f(n))$ so $f(f(n))=k$ for some $k$ in $\{1,2,...,n\}$ and if $k \ne n$ then it will be in contradiction with $f(n) \ge n$ so $f(f(n))=n \implies f(n)=n$

Solution