Let $A : \mathbb{R}^n \to \mathbb{R}^n$ be an orthogonal linear map with determinant $1$. Show that $$A^* \circ \star = \star \circ A^* : \mathrm{Alt}^k(\mathbb{R}^n) \to \mathrm{Alt}^{n-k}(\mathbb{R}^n)$$ for $0 \le k \le n.$
My working so far. Let $\{e_1, \dots, e_n\}$ be an orthogonal base for $\mathbb{R}^n$. We then get an orthogonal base $\{\varepsilon_{i_1} \wedge \dots \wedge\varepsilon_{i_k} \mid 1\le i_1 < \dots < i_k \le n \}$ for $\mathrm{Alt}^k(\mathbb{R}^n)$. Now since $A: \mathbb{R}^n \to \mathbb{R}^n$ we know that $A^* : \mathrm{Alt}^k(\mathbb{R}^n) \to \mathrm{Alt}^k(\mathbb{R}^n)$ is a multiplication by $\det(A)$. For $v_1,\dots,v_k \in \mathbb{R}^n$ we have $$\begin{align*}A^*(\star(\varepsilon_{i_1} \wedge \dots \wedge\varepsilon_{i_k}))(v_1,\dots,v_k)&= A^*(\varepsilon_{i_{k+1}} \wedge \dots \wedge\varepsilon_{i_{n}})(v_1,\dots,v_k) \\ &= \det(A)\varepsilon_{i_{k+1}} \wedge \dots \wedge\varepsilon_{i_{n}}(A(e_1),\dots,A(e_k)) \\ &= \varepsilon_{i_{k+1}} \wedge \dots \wedge\varepsilon_{i_{n}}(A(e_1),\dots,A(e_k)).\end{align*}$$
Similarly $$\begin{align*} \star(A^*(\varepsilon_{i_1} \wedge \dots \wedge\varepsilon_{i_k}))(v_1,\dots,v_k) &= \star(\det(A)\varepsilon_{i_1} \wedge \dots \wedge\varepsilon_{i_k}(A(e_1),\dots,A(e_k))) \\&= \star(\varepsilon_{i_1} \wedge \dots \wedge\varepsilon_{i_k}(A(e_1),\dots, A(e_k)) \\&= \varepsilon_{i_{k+1}} \wedge \dots \wedge\varepsilon_{i_n}(A(e_1),\dots,A(e_k))\end{align*}$$
which would imply equality. My issue is two things. Firstly I'm not using orthogonality at all here which probably means I'm doing something wrong. Second issue is that is it enough to show that the equality holds for basis elements $\varepsilon_{i_1} \wedge \dots \wedge\varepsilon_{i_k}$ and argue that linearity will give the general case?