exterior derivative for parallelizable manifolds Suppose M has tangent bundle $TM = V \times M$, where $V$ is a vector space.
Then all exterior bundles have the form $\wedge^p (T^*M) = \wedge^p V \times M $.
In particular, $p$-forms on $M$ are the same as functions $\alpha\colon M \to \wedge^p V^*$.
As a consequence, the $p+1$-form $d\alpha$ will correspond to a function $g\colon M \to \wedge^{p+1} V^*$.
What is a formula for $g$?
Edit: the question might be harder that I thought. Let's assume $M=G$ is a Lie group and that the trivialization of $\wedge^n T^*G$ is obtained by left translating a non-zero element of $\mathfrak g = T_e G$. Do we have a nice formula in this case?
 A: Let $X_1,..X_n$ be a trivialisation of the tangent bundle, f and $[X_i,X_j]= \sum _k c_{ij}^k X_k$.Then one can describe the exterior derivative in tremes of the fubnctions $c_{ij}^k$.
If $\omega ^i$ is the 1-form defined by $\omega ^i (X_j)= \delta ^i_j$, we derive and obtain :
$d\omega ^i (X_k,X_k)= \omega ^i ([X_k,X_l])= c^i _{k,l}$
or $(*) \hskip 20pt d\omega ^i = \sum _{k,l} c^i _{k,l} \omega ^k \wedge \omega ^l$
A general form is $\omega = \sum _i a_i \omega ^i$, so that 
$(**) \hskip 20 pt d \omega = \sum _{i,j} X_j(a_i) \omega ^j \wedge \omega ^i + \sum _{k,l} a_i c^i _{k,l} \omega ^k \wedge \omega ^l$
where $X_j(a_i)$ is the derivative $<X_j, da_j>$.
Now we can describe the exterior derivative for a form of any degree by using the product formula $d  f\omega = df \wedge \omega$ and $d (\omega _1 \wedge \omega _2 )= d (\omega _1 )\wedge \omega _2 + (-)^{deg \omega _1} \omega _1 \wedge d \omega _2$. 
The general formula is easily computed but not so nice as the formula $*$. 
Remark 1 : The case where the functions $c_{i,j}^k$ are constant is excatly the case of a (local) Lie group.
Remark 2. The systematic use of Einstein's convention   simplifies the calculations, for instance $
**$ reads
$d \omega = da_i\omega ^i=   X_j(a_i) \omega ^j \wedge \omega ^i +   a_i c^i _{k,l} \omega ^k \wedge \omega ^l$ 
