There are various equivalent ways of introducing the wedge product. One possibility is to first define $\,\wedge\,$ for basis forms, $$ e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_r}\;\equiv\;r!\,\left[\,e^{i_1}\,\otimes\;.\,.\,.\;\otimes\,e^{i_r}\,\right] \quad,\qquad $$ (where [...] = Alt = $\frac{\textstyle 1}{\textstyle k!}\sum \,.\,.\,.\,$), and then to employ the simple theorem $$ \left[\;\; \left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_k}\,\right]\;\otimes\;\left[\,e^{j_{k+1}}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right] \;\;\right] \;=\;\left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right] $$ as a means to extend the definition of $\,\wedge\,$ to arbitrary skew forms.
To this end, we rewrite the above equality as $$ \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\;=\; \frac{\textstyle 1}{\textstyle (k+l)!}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} ~\qquad $$ and then as $$ \frac{(k+l)!}{k!\;l!}\;\left[\;\;\left(\,\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\;=\; e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} ~\;\;.\qquad\qquad $$ We now see that, if we extend the definition of $\,\wedge\,$ to arbitrary exterior forms as $$ \omega^k\,\wedge\,\omega^l\;\equiv\;\frac{(k+l)!}{k!\;l!}\;\left[\;\omega^k\,\otimes\,\omega^l\;\right]\;\; $$ and apply it to $\;\left(\,\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\;$ and $\;\left(\,\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;$, we end up with $$ \left(\,\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\wedge\, \left(\,\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;=\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}}\;\,. $$ This serves as a motivation to introduce the factor of $\,{(k+l)!}/(k!\,l!)\,$ in the above definition. The purpose of these factorials is to make the operation $\,\wedge\,$ associative.
Consider three forms of the orders $\,k\,$, $\,l\,$, and $\,k+l\;$: $$ \alpha^k\;=\;\frac{1}{k!}\;\sum_{i_1\,.\,.\,.\,i_k}^n\,\alpha_{i_1\;.\;.\;.\;i_k}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \;=\;\sum_{i_1\,<\,.\,.\,.\,<\,i_k}^n \,\alpha_{i_1\;.\;.\;.\;i_k}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \quad, $$ $$ \beta^l\;=\;\frac{1}{l!}\;\sum_{i_{k+1}\,.\,.\,.\,i_{k+l}}^n\,\beta_{i_1\;.\;.\;.\;i_l}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \;=\;\sum_{i_{k+1}\,<\,.\,.\,.\,<\,i_{k+l}}^n\,\beta_{i_1\;.\;.\;.\;i_l}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}}\quad, $$ $$ \omega^{k+l}=\frac{1}{(k+l)!}\sum_{i_1\,.\,.\,.\,i_{k+l}}^n\,\omega_{i_1\;.\;.\;.\;i_{k+l}}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k}\;=\sum_{i_1\,<\,.\,.\,.\,<\,i_{k+l}}^n\,\omega_{i_1\;.\;.\;.\;i_{k+l}}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \;\;. $$
Now, the Question:
If we assume that $$ \omega^{k+l}\,=\,\alpha^k\wedge\beta^l\;\;, $$ how to employ the definition $$ \omega^{k+l}\,=\;\alpha^k\wedge\beta^l\,\equiv\;\frac{(k+l)!}{k!\,\;l!}\;\left[\,\alpha^k\,\otimes\,\beta^l\,\right] $$ to express the components $\;\omega_{i_1\;.\;.\;.\;i_{k+l}}\;$ via $\;\alpha_{i_1\;.\;.\;.\;i_k}\;$ and $\;\beta_{i_1\;.\;.\;.\;i_l}\;\;?$
My first attempt was to start out with $$ \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\;=\; \frac{\textstyle 1}{\textstyle (k+l)!}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} ~\;\;, $$ multiply both sides with some $\;\alpha_{i_1\;.\;.\;.\;i_k}\;\,\beta_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;$: $$ \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,\alpha_{i_1\;.\;.\;.\;i_k}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,\,\beta_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right] $$ $$ =\;\frac{\textstyle 1}{\textstyle (k+l)!}\;\;\alpha_{i_1\;.\;.\;.\;i_k}\;\,\beta_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \qquad $$ and sum over all indices on both sides. Comparing the outcome with the definition $$ \left[\;\alpha^k\,\otimes\,\beta^l\;\right]\;=\; \frac{k!\;l!}{(k+l)!}\; \alpha^k\wedge\beta^l\;\;, $$ thus I arrived at $$ \omega_{j_1\;.\;.\;.\;j_{k+l}}\;\equiv\;\frac{(k+l)!}{k!\;l!}\;\alpha_{j_1\;.\;.\;.\;j_k}\;\beta_{j_{k+1}\;.\;.\;,\;j_{k+l}}\;\;.\qquad $$ This result has the correct number of terms, $\;{(k+l)!}/({k!\;l!})\;$. However, all of these come out equal, while in reality they must contain the alternating factor, and the right answer must be $$ \omega_{j_1\;.\;.\;.\;j_{k+l}}\;\equiv\;\sum_{i_1\;.\,.\,.\;i_k}\sum_{i_{k+1}\,.\,.\,.\,i_{k+l}}\delta^{i_1\;.\,.\,.\;i_{k+l}}_{j_1\;.\;.\;.\;j_{k+l}}\; \alpha_{i_1\;.\;.\;.\;i_k}\;\beta_{i_{k+1}\;.\;.\;,\;i_{k+l}}\;\;.\qquad $$ It is not, however, clear to me how to derive it.