Let function $g:A\to A$, $\:\:\:g^{2016}=g\circ g\circ g...\circ g=id_A$ , Prove that $g$ is Inverse function
My attempted :
$g$ is Inverse function if it's Bijection
to prove $g$ is one-to-one function we need to show that $g$ is Left inverses :
we get that :$$g^{2015}(g(a))=id_A$$
so $g$ is one-to-one function
and to prove $g$ is Surjective function we need to show that $g$ is right inverses :
we get that :$$g(g^{2015}(a))=id_A$$
so $g$ is Surjective function
$\implies$ $g$ is Bijection so it's Inverse function.
Is this answer is correct and enough to prove it , if not how to prove it the right way ?
thanks a lot.