The second edition of Herstein's Topics in Algebra (pg 62) has a proof for Sylow's theorem for abelian groups.
In this he mentions that since $p| O(G/S) $ for some subgroup $S$ of the abelian group $G$, there is an element $Sx$ in $G/S$ satisfying $Sx \neq S $ and $(Sx)^{p^ {n}} = S$ for some integer $n>0$.
I do not understand why $(Sx)^{p^ {n}} = S$.
Just before this theorem he proved Cauchy's theorem for abelian groups. So I get that $(Sx)^{p} = S$ is true for some element $Sx$. But why is there a need to generalise for $p^n$ for $n>0$.
Note: In this proof $S = \{ x \in G \ | x^{p^{n}} = e\ for \ some\ integer\ n\}$
here is the complete proof (pg 62).
The quotient $G/S$ is an abelian group whose order is divisible by $p$. Apply Cauchy's theorem to it.
There exists an element $Sx \in G/S$ whose order is $p$, that is, $(Sx)^p = S$. He's taking $n$ to be $1$.
I agree with you and don't honestly see why you would need some other $n > 1$. The contradiction in the next paragraph works fine with $n = 1$.
Side note: By induction on $n$, it is true that, if $(Sx)^p = S$, then $(Sx)^{p^n} = S$ for all $n > 0$.