I would like to give some details in order to make clear that one can give a proof with hardly any computations at all (I have never looked at the Bourbaki presentation but I guess they make the same point though, because they want to make all proofs only depend on previous material they might make a few more computations).

To begin with we are considering supercommutative algebras (over a commutative base ring $R$) which are strictly commutative, i.e., $\mathbb Z/2$-graded algebras with $xy=(-1)^{mn}yx$, where $m$ and $n$ are the degrees of $x$ and $y$ <em>and</em> $x^2=0$ if $x$ has odd degree. Note that the base extension of such an algebra is of the same type (the most computational part of such a verification is for the strictness which uses that on odd elements $x^2$ is a quadratic form with trivial associated bilinear form). The exterior algebra is then the free strictly supercommutative algebra on the module $V$. Furthermore, the (graded) tensor product of two sca's is an sca and from that it follows formally that $\Lambda^*\ast(U\bigotimes V)=\Lambda^\ast U\bigotimes\Lambda^\ast V$. 

Now, the diagonal map $U\to U\bigotimes U$ then induces a coproduct on $\Lambda^\ast U$ which by functoriality is cocommutative making the exterior algebra into superbialgebra. I also want to make the remark that if $U$ is f.g. projective then so is $\Lambda^\ast U $. Indeed, by presenting $U$ as a direct factor ina free f.g. module one reduces to the case when $U$ is free and then by the fact that direct sums are taken to tensor products to the case when $U=R$ but in that case it is clear that $\Lambda^\ast U=R\bigoplus R\delta$ where $\delta$ is an 
odd element of square $0$.

If now $U$ is still f.g. projective then $(\Lambda^\ast U^\ast)^\ast$ is also a supercommutative and supercocommutative superbialgebra. We need to know that it is strictly supercommutative. This can be done by a calculation but we want to get by with as much handwav
ing as possible. We can again reduce to the  case when $U$ is free. After that we can reduce to the case when $U=R$ (using that the tensor product of sca's is an sca) and there it is clear (and one can even in that case avoid computations). Another way, once we are then free case, is to use the base change property to reduce to the when $R=\mathbb Z$ and then we have seen that the exterior algebra is torsion free and a torsion free surcommutative algebra is an sca.

We also need to know that if $U$ is f.g. projective, then the structure map $U\to\Lambda^\ast U$ has a canonical splitting. This is most easily seen by introducing a $\mathbb Z$-grading extending the $\mathbb Z/2$-grading and showing that the degree $1$ part of it is exactly $U$ (maybe better would have to work from the start with $\mathbb Z$-gradings).

This splitting gives us a map $U\to(\Lambda^\ast U^\ast)^\ast$ into the odd part and as we have an sca we get a map $\Lambda^\ast U\to(\Lambda^\ast U^\ast)^\ast$ of sca's. This map is an isomorphism. Indeed, we first reduce to the case when $R$ is free (this is a little bit tricky if one wants to present $U$ as a direct factor of free but one may also use base change compatibility to reduce to the case when $R$ is local in which case $U$ is always free). Then as before one reduces to $U=R$ where itt is clear as both sides are concentrated in degrees $0$ and $1$ and the map clearly is an iso in degree $0$ and by construction in degree $1$.

Note that everything works exactly the same if one works with ordinary cmmutativity (so that the exterior algebra is replaced by the symmetric one) up till the verification in the case when $U=R$. In that case the dual algebra is the divided power algebra and we only have an isomorphism in characteristic $0$.