This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group dealing with Barsotti-Tate groups and here I would like to clarify a proof presented by Anonymous in the comments from linked thread which I not completly understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.

Assume $G$ is a finite flat commutative group scheme over a field $k$ of order $p^N$. Assume $p$ prime and $p \in k \backslash \{0\}$, equivalently invertible on the base.

Claim: Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base.

Anonymous' proof works as follows: Firstly we reduce to case over a field (because a finitely presented flat map is etale if it is fibrewise etale). Since we assumed $G$ commutative the multiplication by $p^N$ map $f_{p^N}: G \to G$ is well defined and by Deligne's theorem $p^N$ kills $G$ since it's the order of $G$. That means that $f_{p^N}$ is the zero map: equivalently it factorize over $Spec(k)$.

What now comes I not understand:

It is clamed that $f_{p^N}$ is unramified "as the map on the tangent spaces is given by $p^N$, which is invertible".

Question I: why the induced by $f_{p^N}$ maps on tangent spaces is given by $p^N$?

Question II: assume we understand Question 1. Why this implies $G$ unramified?

when we can answer these two question we are done because unramified finite type schemes over a field are etale.

This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group dealing with Barsotti–Tate groups and here I would like to clarify a proof presented by Anonymous in the comments from linked thread which I do not completely understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.

Assume $G$ is a finite flat commutative group scheme over a field $k$ of order $p^N$. Assume $p$ prime and $p \in k \setminus \{0\}$, equivalently invertible on the base.

Claim: Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base.

Anonymous' proof works as follows: Firstly we reduce to case over a field (because a finitely presented flat map is etale if it is fibrewise etale). Since we assumed $G$ commutative the multiplication by $p^N$ map $f_{p^N}: G \to G$ is well defined and by Deligne's theorem $p^N$ kills $G$ since it's the order of $G$. That means that $f_{p^N}$ is the zero map: equivalently it factorizes over $\operatorname{Spec}(k)$.

What now comes I do not understand:

It is claimed that $f_{p^N}$ is unramified "as the map on the tangent spaces is given by $p^N$, which is invertible".

Question I: why is the induced by $f_{p^N}$ maps on tangent spaces given by $p^N$?

Question II: assume we understand Question I. Why does this imply $G$ is unramified?

when we can answer these two question we are done because unramified finite type schemes over a field are etale.

This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group dealing with Barsotti-Tate groups and here I would like to clarify a proof presented by Anonymous in the comments from linked thread which I not completly understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.

Assume $G$ is a finite flat commutative group scheme over a field $k$ of order $p^N$. Assume $p$ prime and $p \in k \backslash \{0\}$, equivalently invertible on the base.

Claim: Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base.

Anonymous' proof works as follows: Firstly we reduce to case over a field (because a finitely presented flat map is etale if it is fibrewise etale). Since we assumed $G$ commutative the multiplication by $p^N$ map $f_{p^N}: G \to G$ is well defined and by Deligne's theorem $p^N$ kills $G$ since it's the order of $G$. That means that $f_{p^N}$ is the zero map: equivalently it factorize over $Spec(k)$.

What now comes I not understand:

It is clamed that $f_{p^N}$ is unramified "as the map on the tangent spaces is given by $p^N$, which is invertible".

Question I: why the induced by $f_{p^N}$ maps on tangent spaces is given by $p^N$?

Question II: assume we understand Question 1. Why this implies $G$ unramified?

when we know this we are done because unramified finite type schemes over a field are etale.

This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group dealing with Barsotti-Tate groups and here I would like to clarify a proof presented by Anonymous in the comments from linked thread which I not completly understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.

Assume $G$ is a finite flat commutative group scheme over a field $k$ of order $p^N$. Assume $p$ prime and $p \in k \backslash \{0\}$, equivalently invertible on the base.

Claim: Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base.

Anonymous' proof works as follows: Firstly we reduce to case over a field (because a finitely presented flat map is etale if it is fibrewise etale). Since we assumed $G$ commutative the multiplication by $p^N$ map $f_{p^N}: G \to G$ is well defined and by Deligne's theorem $p^N$ kills $G$ since it's the order of $G$. That means that $f_{p^N}$ is the zero map: equivalently it factorize over $Spec(k)$.

What now comes I not understand:

It is clamed that $f_{p^N}$ is unramified "as the map on the tangent spaces is given by $p^N$, which is invertible".

Question I: why the induced by $f_{p^N}$ maps on tangent spaces is given by $p^N$?

Question II: assume we understand Question 1. Why this implies $G$ unramified?

when we can answer these two question we are done because unramified finite type schemes over a field are etale.