Let $V$ denote an $n$-dimensional real vector space, and $\wedge^n$ denote the $n$-fold exterior product. What is the isomorphism between $\wedge^n(V)$ and $\mathbb{R}$?
In the book Introduction to Manifolds, by Loring W. Tu, 3rd edition, Chapter: Orientation of Manifolds, it says:
A linear isomorphism $\wedge^n(V)\cong\mathbb{R}$, identifies the set of non-zero $n$-covectors on $V$ with $\mathbb{R} \setminus \{0\}$".
Note that you assume that $V$ is of dimension $n$. The $n$-fold outer product of $V$ can now be described as follows: Choose a basis $v_1,\dots,v_n$ of $V$. The outer product is generated (as a vector space) by expressions of the form $v^1\wedge\dots\wedge v^n$ where $v^1,\dots,v^n\in\{v_1,\dots,v_n\}$, where one such expression $v^1\wedge\dots\wedge v^n$ is the negative of $w^1\wedge\dots\wedge w^n$ if $w^1\wedge\dots\wedge w^n$ is obtained from $v^1\wedge\dots\wedge v^n$ by exchanging one component $v^i$ and another component $v^j$. This implies that $v^1\wedge\dots\wedge v^n=0$ if there are distinct $i$ and $j$ with $v^i=v^j$.
It follows that $v_1\wedge\dots\wedge v_n$ and $-v_1\wedge\dots\wedge v_n$ are the only non-zero generators, and they are linearly dependent (obviously). Hence the outer product is generated by $v_1\wedge\dots\wedge v_n$. Mapping this generator to $1$ extends to your desired isomorphism. As pointed out in the comments, there are many isomorphisms, and the one pointed out here depends on the choice of the basis $v_1,\dots,v_n$.