Distributivity of wedge product over addition The book I'm reading uses the definition of wedge product of n one-forms over $v_n$ vectors, by the determinant formula:
$$
d x_{i_1} \wedge d x_{i_2} \wedge \cdots \wedge d x_{i_n}\left(v_1, v_2, \ldots v_n\right)=\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{j=1}^n d x_{\sigma\left(i_j\right)}\left(v_j\right)
$$
With $a_1, a_2, b \in \mathbb{R}$, let $\omega_1=a_1 d x_{i_1} \wedge \cdots d x_{i_k}, \omega_2=a_2 d x_{j_1} \wedge \cdots d x_{j_k}$, and $\eta=b d x_{\ell}$. Use the formula above to show the distributivity property
$$
\left(\omega_1+\omega_2\right) \wedge \eta=\omega_1 \wedge \eta+\omega_2 \wedge \eta
$$
Every other book that I found just assumed this property is true or left as an exercise to the reader. I tried expanding $\omega_1$ and $\omega_2$ using the summation, though I got nowhere. How can I prove this?
 A: In that case, it is easier to write the exterior product as an antisymmetrized tensorial product, in order to avoid the reference to the exterior algebra basis, which tends to make the computations muddled.
Concretely, let $\omega_1,\omega_2 \in \Omega^k(V)$, $\eta \in \Omega^l(V)$ be $k$- and $l$-forms over space $V$ respectively and $v_1,\ldots,v_{k+l} \in V$ some vectors. Then we have :
$$
\begin{array}{l}
((\omega_1+\omega_2)\wedge\eta)(v_1,\ldots,v_{k+l}) \displaystyle \color{white}{\frac{1}{1}} \\
\begin{array}{rcl}
\color{white}{dd}
   &=& \displaystyle
   \mathrm{Alt}((\omega_1+\omega_2)\otimes\eta)(v_1,\ldots,v_{k+l}) \\
   &=& \displaystyle
   \frac{1}{(k+l)!}\sum_{\sigma\in S_{k+l}}\mathrm{sgn}(\sigma) ((\omega_1+\omega_2)\otimes\eta)(v_{\sigma(1)},\ldots,v_{\sigma(k+l)}) \\
   &=& \displaystyle
   \frac{1}{(k+l)!}\sum_{\sigma\in S_{k+l}}\mathrm{sgn}(\sigma) (\omega_1+\omega_2)(v_{\sigma(1)},\ldots,v_{\sigma(k)}) \cdot \eta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)}) \\
   &=& \displaystyle
   \frac{1}{(k+l)!}\sum_{\sigma\in S_{k+l}}\mathrm{sgn}(\sigma) \left\{\omega_1(v_{\sigma(1)},\ldots,v_{\sigma(k)})\eta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)}) + \omega_2(v_{\sigma(1)},\ldots,v_{\sigma(k)})\eta(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)})\right\} \\
   &=& \displaystyle
   \frac{1}{(k+l)!}\sum_{\sigma\in S_{k+l}}\mathrm{sgn}(\sigma) (\omega_1\otimes\eta + \omega_2\otimes\eta)(v_{\sigma(1)},\ldots,v_{\sigma(k+l)}) \\
   &=& \displaystyle
   \mathrm{Alt}(\omega_1\otimes\eta + \omega_2\otimes\eta)(v_1,\ldots,v_{k+l}) \\
   &=& \displaystyle
   (\omega_1\wedge\eta + \omega_2\wedge\eta)(v_1,\ldots,v_{k+l}) \color{white}{\frac{1}{1}}
\end{array}
\end{array}
$$
where we used only the definitions of the different objects.
