In Fulton & Harris's "Representation Theory: A First Course" it is claimed that the wedge product $\wedge$ determines a product $\text{Alt}^m V \times \text{Alt}^n V \to \text{Alt}^{m+n} V$, defined by $$(v_1 \wedge \dots \wedge v_m, v_{m+1} \wedge \dots \wedge v_{m+n}) \mapsto v_1 \wedge v_2 \wedge \dots \wedge v_{m+n}$$ Why is that the case? How would one show that this map is well defined?
One way that that would most likely work is embedding the alternating powers in the tensor powers and verifying it by hand, but I assume there has to be a less grueling way, mayhap using the universal property of the alternating powers.
Thank you in advance
$ \newcommand\Alt{\mathop{\mathrm{Alt}}} $Consider the map $\psi : V^{m+n} \to \Alt^{m+n}V$ defined by $$ \psi(v_1,\dotsc,v_{m+n}) = v_1\wedge\dotsb\wedge v_{m+n}. $$ Fix $v_{m+1},\dotsc,v_{m+n}$ so that we have a map $V^m \to \Alt^{m+n}V$. It should be evident that this is alternating, so by the universal property of $\Alt^mV$ it extends uniquely to a map $\Alt^mV \to \Alt^{m+n}V$. Letting $v_{m+1},\dotsc,v_{m+n}$ vary we now have a map $\psi' : \Alt^mV\times V^n \to \Alt^{m+n}$ such that $$ \psi'(v_1\wedge\dotsb\wedge v_m, v_{m+1},\dotsc,v_{m+n}) = v_1\wedge\dotsb\wedge v_{m+n}. $$ Now for fixed $X \in \Alt^mV$ consider the map $V^n \to \Alt^{m+n}V$ given by $$ (v_{m+1},\dotsc,v_{m+n}) \mapsto \psi'(X, v_{m+1}, \dotsc, v_{m+n}). $$ This is evidently alternating and so extends uniquely to a map $\Alt^nV \to \Alt^{m+n}V$, and by letting $X$ vary we finally have a map $\Alt^mV\times\Alt^nV \to \Alt^{m+n}V$ satisfying $$ (v_1\wedge\dotsb\wedge v_m,\: v_{m+1}\wedge\dotsb\wedge v_{m+n}) \mapsto v_1\wedge\dotsb\wedge v_{m+n}. $$