Is it true that $\Gamma(\Lambda(T^*M)) \cong \Lambda(\Gamma(T^*M))$? Well, the question is in the title. 
Is it true, that given a smooth manifold $M$, the following isomorphism holds:
$$
\Gamma(\Lambda(T^*M)) \cong \Lambda(\Gamma(T^*M))
$$
$\Gamma$ - smooth sections functor, $\Lambda$ - exterior algebra functor. Base ring for both - $C^\infty(M)$.
Motivation: I'm kind of confused, because some sources define differential forms as sections of exterior bundle, others define forms as elements of exerior algebra of sections:
For example:


*

*Wikipedia uses first definition https://en.wikipedia.org/wiki/Differential_form

*nLab uses second definition https://ncatlab.org/nlab/show/exterior+algebra (at the bottom)


Are those equivalent?
 A: Working with smooth manifolds, these are equivalent notions because you can use partitions of unity to extend local $k$-forms to be global ones, thus writing a global $k$-form as the sum of the wedge products of global $1$-forms.
The equivalence is false in the holomorphic category, not surprisingly. The simplest example I can think of is a smooth quartic ($K3$) surface $X\subset\Bbb CP^3$. It has trivial canonical bundle, so $\Lambda^2 T^*X$ is the trivial bundle, hence has global sections. On the other hand, $T^*X$ has no global holomorphic sections other than $0$ because  $h^0(X,\Omega^1_X) = h^{1,0}(X) = h^{0,1}(X) = h^1(X,\mathscr O_X) = 0$.
A: By definition, a differential form of order $k$ is nothing but a section of $\Lambda^k \left(T^{*} M\right).$ It is natural unlike $\Gamma\left(\Lambda\left(T^{*} M\right)\right)$, because you want differential forms to form a bundle, so any differential form should be a section of some bundle.  In particular, a differential forms of order 1 are a space $\Gamma(T^{*} M).$  Now let us describe $\Lambda^k \Gamma(T^{*} M)$, namely: since any 1-differential form has a local basis $dx_{i},$ then any element of $\Lambda^k \Gamma(T^{*} M)$ has a local basis $dx_{i_1} \wedge .. \wedge dx_{i_k}$, therefore this is a space of $k$-differential forms. Thus, you have an identification between these two spaces.
