I want to show, that the $\wedge$-product is bilinear.
It is $\displaystyle (\omega^k\wedge\eta^l)(v_1,\dotso,v_{k+l}):=\frac{1}{k!l!} \sum_{\sigma\in S_{k+l}} \operatorname{sgn} (\sigma) \omega^k (v_1, \dotso, v_k) \eta(v_{k+1},\dotso,v_{k+l})$
I have to show, that for forms of appropriate degree it is:
1) $(\omega_1^k + \omega_2^k) \wedge \eta^l = \omega^k_1 \wedge \eta^l + \omega_2^k \wedge\eta^l$
2) $\omega^k \wedge (\eta_1^l+\eta_2^l) = \omega^k \wedge \eta_1^l + \omega^k \wedge \eta_2^l$
3)$(\alpha\omega^k)\wedge\eta^l=\omega^k\wedge(\alpha\eta^l)=\alpha(\omega^k\wedge\eta^l)$ with $\alpha\in\mathbb{K}$
3) is completly trivial and 2) should be the same as 1). So I have a question to 1). My proof goes as follows:
$$(\omega_1^k+\omega_2^k)\wedge\eta^l=\frac{1}{k!l!}\sum_{\sigma\in S_{k+l}} \operatorname{sgn}(\sigma)(\omega_1^k+\omega_2^k)(v_1,\dotso, v_k)\eta^l(v_{k+1},\dotso,v_{k+l})$$
The forms are already multilinear. Therefore we have
$$(\omega_1^k+\omega_2^k)(v_1,\dotso, v_k) = \omega_1^k (v_1, \dotso, v_k) + \omega_2^k (v_1,\dotso,v_k)$$
and it what we have to show becomes trivial. But I am unsure if the step above is true.
Thanks in advance for clarification.
The wedge product is defined as follows. Take two forms $\omega,\eta$. Their tensor product is the form $\omega\otimes \eta$ such that (for tuples $v,w$ of vectors of the correct length)
$$(\omega\otimes \eta)(v,w) = \omega(v)\eta(w)$$
If $\omega$ is a form of degree $k$ and $\sigma$ is a permutation of degree $k$ there is a another form $\sigma\cdot \omega$ so that $\sigma\cdot \omega(w) = \omega(\sigma^{-1}w)$ where $\sigma^{-1}=\tau$ acts on $w$ by shuffling its coordinates according to the order dictated by $\tau$.
Now $\omega\wedge \eta$ is obtained from $\omega\otimes\eta$ by applying the element $$\varepsilon = \frac{1}{p!q!}\sum_{\sigma\in S_{p+q}}(-1)^\sigma\sigma$$ which behaves linearly with respect to sums of tensors. Thus the fact that the wedge product is bilinear follows from the fact the tensor product of forms is bilinear.
For example, it is evident from the definition of the tensor product that $$(\omega_1+\omega_2)\otimes \eta = \omega_1\otimes \eta+\omega_2 \otimes \eta,$$
and applying the linear transformation $\varepsilon$ gives that $$(\omega_1+\omega_2)\wedge \eta = \omega_1\wedge \eta+\omega_2 \wedge \eta.$$