I have been reading this [article](https://reader.elsevier.com/reader/sd/pii/S0001870809000450?token=B3552A852B4AF2746FAC58DF2245E2AD016AA6D5C38807AA034E0E48ACA1FC4F606A1D82B733DF71E58FC1E68F56892E&originRegion=eu-west-1&originCreation=20210509103957). In the Example 2.17, page 1155, there is a claim that for $m>1$ and $n\geq 1$, $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$.

Let me explain the notation quickly: 

  $\mu_n$ is the group scheme of $n^{th}$ roots of unity acting on $\mathbb{A}^m-0$ by diagonal action. So $\mathbb{A}^m-0/\mu_n$ is a smooth scheme. $\mathbb{G}_m/\mathbb{G}_m^n$ is the cokernel of the $n^{th}$-power map $\mathbb{G}_m\xrightarrow{(-)^n} \mathbb{G}_m$ in the category of Nisnevich sheaves of abelian groups. 

My question is how to see the isomorphism  $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$? 

Comments are most welcome!