Multiplication by n on an abelian scheme is finite flat, and etale for n invertible

By [F1] every geometric fibre of $[n]:A\to A$ has finite kernel of order $n^{2g}$, so $[n]$ is quasi-finite; it is proper because $A$ is proper over $S$, hence finite by [F2]. Consequently $A[n]=\ker [n]$ is finite over $S$ and of finite type.

Flatness of $[n]$ follows by applying the Noetherian flatness-by-fibres criterion [F2] to the local tower ${\mathcal{O}}_{S,s}\to {\mathcal{O}}_{A,y}\to {\mathcal{O}}_{A,x}$ for $[n]$, with $M={\mathcal{O}}_{A,x}$: smoothness makes $M$ flat over ${\mathcal{O}}_{S,s}$, and the field-level multiplication theorem makes the special-fibre module flat over the special-fibre target. Constancy of the rank then follows from the rank $n^{2g}$ on geometric fibres, so $[n]$ is finite flat of degree $n^{2g}$ and $A[n]$ has rank $n^{2g}$.

If $n$ is invertible on $S$, then the differential of $[n]$ at the identity is multiplication by the unit $n$ on the locally free sheaf of invariant differentials (the differential of the group law is addition), and translation-equivariance spreads this to every point; the equal-relative-dimension criterion of [F2] therefore makes $[n]$ etale, and $A[n]\to S$, being the pullback along the identity section, is finite etale of rank $n^{2g}$.