Let $V$ be an $n$-dimensional real inner product space. An isometry on $V$ is an operator $R$ with $\langle Rx,Ry \rangle$ for all $x,y \in V$. The determinant of an Endomorphism can be defined in many ways. One of them is as follows. Since the the space $\Lambda^n V$ is $1$-dimensional then for any basis $\{v_j\}$ we have $$ Rv_1\wedge\dots\wedge Rv_n = (\det R)\ v_1\wedge\dots\wedge v_n, $$ where $\det R$ is the proportionality factor. Now to show from these definitions that for an isometry $R$ we have $\det R= \pm 1$, I define the inner product on $\Lambda^n V$ such that $$ \langle u_1\wedge\dots\wedge u_n,v_1\wedge\dots\wedge v_n \rangle := \det \langle u_i, v_j \rangle = \sum_{\pi \in S_n} \mathrm{sgn}(\pi) \langle u_1,v_{\pi 1} \rangle \cdots \langle u_n,v_{\pi n} \rangle, $$ and show that $$ \|Rv_1\wedge\dots\wedge Rv_n\| = \| v_1\wedge\dots\wedge v_n \| $$ and hence $|\det R| = 1$. But this might be an exotic way, since it involves a new space with an inner product, but I'm very sure why it works.
Is there another way without any standard basis of $\mathbb R^n$ or any matrices ? In the answer here it relies on the fact $\det R = \det R^T$, where $R^T$ is the adjoint of $R$, but that is not proved in that answer, clearly it is a matrix trick.
By the way one can define $\det R$ as proportionality factor of elements of $\Lambda^n V^*$ as $$ \omega(Rv_1,\dots,Rv_n) = \det R\ \omega(v_1,\dots,v_n) $$ but I couldn't continue from there.