Pick a point $x\in X$. It will be enough to show that there is a neighbourhood
of $x$ on which $\mathcal F$ is free of rank $n$. Now, the exterior power
commutes with pullbacks (aka scalar extensions) so that in particular the fibre
(in the sense of pullback to $\mathrm{Spec}k(x)$) of $\Lambda^m\mathcal F$ at
$x$ equals $\Lambda^m\mathcal{F}_x$. This shows that $\mathcal{F}_x$ is an
$n$-dimensional vector space. After possibly shrinking $X$ we may assume that
there is a an $\mathcal{O}_X$-map $f\colon \mathcal{O}_X^n\to \mathcal F$ which induces
an isomorphism on fibres at $x$. Thus $\Lambda^nf$ is a map between locally free
modules (of rank $1$) that gives an isomorphism on fibres at $x$ and hence is an
isomorphism in a neighbourhood of $x$ so that we may assume that it is a global
isomorphism. The wedge product induces pairings
$\mathcal{F}\times\Lambda^{n-1}\mathcal{F}\to \Lambda^{n}\mathcal{F}$ and
$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to
\Lambda^{n}\mathcal{O}_X^n$, the latter being a perfect pairing. Composing the
second with $\Lambda^nf$ gives a pairing
$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to \Lambda^{n}\mathcal{F}$.
As $\Lambda^\ast f$ is multiplicative we get that the composite
$$\mathcal{O}_X^n\xrightarrow{f}\mathcal{F}\to
\mathrm{Hom}(\Lambda^{n-1}F,\Lambda^{n}\mathcal{F})\xrightarrow{\Lambda^{n-1}f^*}\mathrm{Hom}(\Lambda^{n-1}\mathcal{O}_X^n,\Lambda^{n}\mathcal{F})$$
equals the map induced by the pairing for $\mathcal{O}_X^n$. This is an
isomorphism (as $\mathcal{O}_X^n$ is free of rank $n$ and $\Lambda^nf$ is an
isomorphism) so we get that $f$ is split injective and we may write
$\mathcal{F}$ as `\mathcal{O}_X^n\bigoplus \mathcal G$ for some quasi-coherent sheaf
$\mathcal{G}$. Now, $\Lambda^n(\mathcal{O}_X^n\bigoplus \mathcal{G})$ splits up
as
$$
\bigoplus_{i+j=n}\Lambda^i\mathcal{O}_X^n\bigotimes \Lambda^j\mathcal{G}
$$
and $\Lambda^nf$ is the inclusion into the $j=0$ factor. As that inclusion is an
isomorphism, the other factors are zero but $\Lambda^{n-1}\mathcal{O}_X^n\bigotimes
\Lambda^1\mathcal{G}$ has $\mathcal{G}$ as a direct factor and hence $\mathcal{G}=0$.