How to proceed this computation with differential forms? I've been studying Spivak's differential geometry book and he defines the exterior derivative of $\omega \in \Omega^k(M)$ in a coordinate system $(x,U)$ by
$$d\omega = d\omega_{i_1\cdots i_k}\wedge dx^{i_1}\wedge\cdots \wedge dx^{i_k}.$$
He then say we must show this definition is not dependent on the coordinate system and he says one way is to compute directly. He did in another way but I'm trying to show by direct computation. 
What I did was first to simplify the notation denote the derivative of $\omega$ in the $x$ coordinates by $d(\omega,x)$ and on the $y$ coordinates $d(\omega,y)$. We then want to show that $d(\omega, x)=d(\omega,y)$. 
We have then
$$d(\omega,x) = \dfrac{\partial \omega_{i_1\cdots i_k}}{\partial x^\alpha}dx^\alpha \wedge dx^{i_1}\wedge\cdots \wedge dx^{i_k},$$
now by the chain rule and by computing the differentials in terms of $y$ coordinates we have
$$d(\omega,x) = \dfrac{\partial \omega_{i_1\cdots i_k}}{\partial y^\beta}\dfrac{\partial y^\beta}{\partial x^\alpha}dx^\alpha \wedge \dfrac{\partial x^{i_1}}{\partial y^{j_1}}dy^{j_1}\wedge\cdots\wedge \dfrac{\partial x^{i_k}}{\partial y^{j_k}}dy^{j_k},$$
and this is the same as
$$d(\omega, x) = \left(\dfrac{\partial x^{i_1}}{\partial y^{j_1}}\cdots \dfrac{\partial x^{i_k}}{\partial y^{j_k}}\right)\dfrac{\partial \omega_{i_1\cdots i_k}}{\partial y^\beta}dy^\beta \wedge dy^{j_1}\wedge\cdots\wedge dy^{j_k},$$
now I can't get rid of those extra terms on the left and with them the equality doesn't hold. 
How do I proceed with this? I couldn't think of any ways to proceeding any further.
Thanks very much in advance.
 A: Let me write $\alpha = (\alpha_1, \cdots, \alpha_k)$ and $I =(i_1, \cdots, i_k)$. Then using 
$$w = \sum_I w_I dx^{i_1}\wedge \cdots \wedge dx^{i_k} =  \sum_\alpha w_\alpha dy^{\alpha_1}\wedge \cdots \wedge dy^{\alpha_k}$$
and $dy^{\alpha_j} = \frac{\partial y^{\alpha_j}}{\partial x^i} dx^i$, we have 
$$w_I = \sum_{\alpha} w_\alpha  \frac{\partial y^{\alpha_1}}{\partial x^{i_1}}\cdots \frac{\partial y^{\alpha_k}}{\partial x^{i_k}}\ .$$
Now we compute (using the $x$ coordinate) 
$$dw = \sum_{I,j} \frac{\partial w_I}{\partial x^j} dx^j \wedge dx^{i_1}\wedge \cdots \wedge dx^{i_k}$$
$$= \sum_{I,j} \frac{\partial }{\partial x^j}\bigg( \sum_\alpha w_\alpha \frac{\partial y^{\alpha_1}}{\partial x^{i_1}}\cdots \frac{\partial y^{\alpha_k}}{\partial x^{i_k}} \bigg)dx^j \wedge dx^{i_1}\wedge \cdots \wedge dx^{i_k}$$
Now this is a bit messy, but actually we need only to differentiate the first term in the bracket, because 
$$\frac{\partial ^2 y^{\alpha_k}}{\partial x^j \partial x^{i_k}} \text{is symmetric in $j$, $i_k$, and   }\ \ \  dx^j \wedge dx^{i_k} \text{ is antisymmetric in $j$, $i_k$. }$$
So the sum would be zero. Thus 
$$dw = \sum_{\alpha, I,j} \frac{\partial w_\alpha}{\partial x^j} \frac{\partial y^{\alpha_1}}{\partial x^{i_1}}\cdots \frac{\partial y^{\alpha_k}}{\partial x^{i_k}} \ dx^j \wedge dx^{i_1}\wedge \cdots \wedge dx^{i_k} $$
$$= \sum_{\alpha, I, j, \beta} \frac{\partial w_\alpha}{\partial y^\beta}\frac{\partial y^\beta}{\partial x^j} \frac{\partial y^{\alpha_1}}{\partial x^{i_1}}\cdots \frac{\partial y^{\alpha_k}}{\partial x^{i_k}} \ dx^j \wedge dx^{i_1}\wedge \cdots \wedge dx^{i_k}$$
$$= \sum_{\alpha, \beta} \frac{\partial w_\alpha}{\partial y^\beta} dy^\beta \wedge dy^{\alpha_1}\wedge \cdots\wedge dy^{\alpha_k}= \sum_\alpha dw_\alpha \wedge dy^{\alpha_1}\wedge \cdots\wedge dy^{\alpha_k}$$
