With the following convention for the antisymmetric basis of forms:
$\boldsymbol{\tilde\omega}^{[\mu_1}\otimes ...\otimes\boldsymbol{\tilde\omega}^{\mu_p]} = \frac{1}{p!} \Sigma^{p!}_{i=1}(-1)^{\pi(i)}\boldsymbol{\tilde\omega}^{\mu_1}\otimes ... \otimes\boldsymbol{\tilde\omega}^{\mu_p}$
where the sum is over all permuttions, and $\pi(i)$ is zero for even permutations, and one for odd permutations, we can write a p-form as:
$\boldsymbol{\alpha} = \alpha_{\mu_1...\mu_p}\boldsymbol{\tilde\omega}^{[\mu_1}\otimes ...\otimes\boldsymbol{\tilde\omega}^{\mu_p]}$
And the wedge product can be defined as:
$\boldsymbol{\tilde\omega}^{[\mu_1}\otimes ...\otimes\boldsymbol{\tilde\omega}^{\mu_p]}\wedge\boldsymbol{\tilde\omega}^{[\nu_1}\otimes ...\otimes\boldsymbol{\tilde\omega}^{\nu_q]}=\frac{(p+q)!}{p!q!}\boldsymbol{\tilde\omega}^{[\mu_1}\otimes ...\otimes\boldsymbol{\tilde\omega}^{\mu_p}\otimes\boldsymbol{\tilde\omega}^{\nu_1}\otimes ...\otimes\boldsymbol{\tilde\omega}^{\nu_q]}$
All that is fine by me. But how can we using the previous definitions, prove that
$(\boldsymbol{\alpha}\wedge\boldsymbol{\beta})_{\mu_1...\mu_p\mu_{p+1}...\mu_{p+q}}=\frac{(p+q)!}{p!q!}\alpha_{[\mu_1...\mu_p}\beta_{\mu_{p+1}...\mu_{p+q}]}$
I fail to see how, and I would appreciate the help very very much!