This is just expanding on user19475's comment, and is taken entirely from Schoof's notes, but as this question is the first thing to come up when googling, I thought it might be helpful to give my novice's interpretation of the key points of the proof.
First, by changing the base if needed, we are reduced to showing that the $m$th power map is trivial on the $R$ valued points of $G$, where $G=Spec(A)$ is a finite, free, affine group scheme of rank $m$ over $Spec(R)$.
Then, using Cartier duality, we may identify $G(R)\subset A^{\vee}$, since:$$G(R)=\text{Hom}_{R-Alg}(A,R)\subset Hom_{R-Mod}(A,R)=A^{\vee}$$
Further unravelling definitions, the points of $A^{\vee}$ that belong to $G(R)$ are exactly those $x$ which satisfy $$\Delta(x)=x\otimes x$$
For $\Delta$ the comultiplication on $A^{\vee}$.
The key tool now is the norm map, for any finite, free $R$ algebra $S$, we have the norm map $N:S^*\rightarrow R^*$ on units, given by $x\mapsto \det(\mu_x)$ where $\mu_x$ is the $R$ linear map of multiplication by $x$ on $S$ as an free $R$ module.
The core technical part of the proof is that for any $R\rightarrow S$, the norm map from $(A^{\vee}\otimes S)^*\rightarrow (A^{\vee})^*$ takes $G(S)$ into $G(R)$, and when we set $S=A$, that this norm map is invariant under the action of the translation automorphism $\tau(\alpha)$ on $G(A)$ for any $\alpha$ in $G(R)$.
Once we have these two compatibilities, we simply verify that for the element $\text{Id}_A$ in $G(A)$, we have $$N(\text{Id}_A)=N(\tau(\alpha)\text{Id}_A)=N(\alpha)N(\text{Id}_A)=\alpha^m N(\text{Id}_A)$$
Which gives the result.
It appears to me that in the proof, there doesn't seem to be any specific place where commutativity is required, as we are really only looking to prove the assertion at a single element of $G(R)$. The commutativity is important insofar as it provides the core tool used for the proof, the norm map, via the interpretion of points of $G(R)$ as elements of the Cartier dual.
