Suppose $|A|=n,f:A→A$ is injective $\implies∃k∈[n]:f^k(x)=x$

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math.stackexchange.com/questions/3950538/supp … s-exists-k-in-nfkx-x
For fixed $a$ we have a chain $$a,f(a), f^2(a),...f^{n}(a)$$ $n+1$ elements from the set with $n$ elements, thus at least two of them are the same, so there are $0\leq i<j\leq n$ such that $$f^i(a) = f^j(a) $$
so $$f^i(a) = f^i(f^{j-i}(a))$$

thus, since $f^i$ is also injective, we have $$a= f^{j-i}(a)$$ and thus $k:=j-i$ does the work. (as $1≤k≤n$)