$f^m = f^k$ Proofs

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$1$. Prove that for any $f :J_n → J_n $, there exists a positive integer $m$ such that $f^m = f^k$ for some positive integer $k < m$.

$2.$ Let $f :J_n → J_n$ be a function, and let $m$ and $k$ be positive integers such that $f^m = f^k$ and $m > k$. Prove that the restriction of $f$ to $f^k(J_n)$ is a bijective function from $f^k(J_n)$ to $f^k(J_n)$

Note : $J_n$ represents the set consisting of the positive integers from $1$ to $n$. For example $J_5=\{1,2,3,4,5\}$.

For $1$, I know that $f^m = f^{m-k} \circ f^k$. But, I have no idea on how to proceed beyond that. Any help I can get is appreciated.

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There are only finitely many functions from $J_n$ to itself. Therefore, whenever the length of any list of such functions exceeds the total number of such functions, then two members of that list must be equal. For example, there are $256$ functions from $\{1,2,3,4\}$ to itself, so two of the functions $f, f^2, f^3,\ldots,f^{257}$ must be equal to each other.