Let $R$ be a commutative ring, and let $M$ be a free $R$-module with basis $e_1,\ldots,e_n$. If $I=\{i_1,\ldots,i_k\}, i_1<i_2\cdots<i_k$, is a subset of $\{1,\ldots,n\}$, let $e_I=e_{i_1}\wedge\cdots\wedge e_{i_k}$.
It is known that the exterior algebra $\Lambda M$ is free as an $R$-module, with basis $(e_I)_I$.
Most of the proofs I know to get this fact use one of the following approaches:
Prove that $\Lambda^k M$ is free by showing that the family $(e_I)_I$ is linearly independent over $R$, using determinant-like alternating maps + the universal property of $\Lambda^k M$ (like in Keith's Conrad paper)
Prove that $\Lambda M$ is free by exhibiting a basis of the ideal of the tensor algebra $T(M)$ generated by the elements $x\otimes x, x\in M$, using careful inspection (like in these notes)
Use the canonical isomorphism $\Lambda(M\oplus N)\simeq \Lambda M\widehat{\otimes} \Lambda N$
Question. Do you know any proof of the fact that $\Lambda M$ is free whenever $M$ is free, which relies only the universal property of $\Lambda M$ ?
Of course, the third approach uses a result which is proved using the universal property, but this result is quite strong to prove such a thing...
For the record, the universal property i'm talking about is: any map $f:M\to C$ ($C$ is unital associative) such that $f(x)^2=0$ for all $x\in M$ extends in a unique way to a map $\Lambda M \to C$.