I think I have a little confusion with index notation (concerning p-form). For example, do the coordinates on $\mathbb{R}^n$ $\{x^1,x^2, \dots , x^k\}$ and $\{x^{i\_1},x^{i\_2}, \dots , x^{i\_k}\}$ mean the same? Can they be used equivalently (if no, why not?)? 

To illustrate this more claerly: Suppose we have a p-form $\omega$ on $\mathbb{R}^n$. Then I find just expression like $$\omega = \omega\_{{i\_1}, \dots,{i\_p}}dx^{i\_1} \wedge \dots \wedge dx^{i\_p}.$$
Can I also write instead: 
$$\omega = \omega\_{{1}, \dots,{p}}dx^{1} \wedge \dots \wedge dx^{p} ?$$
Is there a difference, and why? I am confused because if you perfom the exterior derivative of $\omega$ (here I use the first representation) according to this formula $$ d\omega = \sum\_{i=1}^n \frac{\partial \omega\_{{i\_1}, \dots,{i\_p}} }{\partial x^i} dx^i \wedge dx^{i\_1} \wedge \dots \wedge dx^{i\_p}, $$ I don't understand how this can be the same as (now I use the second representation)
$$d\omega = \sum\_{i=1}^n \frac{\partial \omega\_{{1}, \dots,{p}} }{\partial x^i} dx^i \wedge dx^{1} \wedge \dots \wedge dx^{p}.$$ 

Performing the derivative for a 1-form on $\mathbb{R}^2, \omega = \omega\_{{i\_1}}dx^{i\_1} $ I get: 
$$ d\omega = \frac{\partial \omega\_{{i\_1}}}{\partial x^1} dx^1 \wedge dx^{i\_1} + \frac{\partial \omega\_{{i\_1}}}{\partial x^2} dx^2 \wedge dx^{i\_1} . $$ If I use the second representation I get: $$ d\omega = \frac{\partial \omega\_{{1}}}{\partial x^1} dx^1 \wedge dx^{1} + \frac{\partial \omega\_{{1}}}{\partial x^2} dx^2 \wedge dx^{1},$$ where $\frac{\partial \omega\_{{1}}}{\partial x^1} dx^1 \wedge dx^{1}$ cancels, and this derivative is indeed right. 

Then, how do I get to this result using the first derivative? Why is $\frac{\partial \omega\_{{i\_1}}}{\partial x^1} dx^1 \wedge dx^{i\_1} $ equal to 0 and $ dx^2 \wedge dx^{i\_1}$ equal to $ dx^2 \wedge dx^{1} $?