Definitions: Define the following map: let $k \in H^m(G_n(\Bbb F^\infty ), R)$, and $\xi= (E,B,p)$ then define $k(\xi) = g^*(k) \in H^m(B,R)$, where $g$ is the induced map from the bundle map $(F,g):\xi \rightarrow \gamma_n$, $\gamma_n$ being the canonical bundle over $G_n(\Bbb F^\infty)$, the Grassmanian.
This is well defined by lemma below.
By lemma again that $k$ induces a natural in the set of $n$-vector bundle isomorphism class. Hence $k$ defines a natural transformation $Vect_n(-)$ to $H^m(-,R)$. This is by defintion, a characteristic class of degree $m$, this set is denoted $\Lambda_m$
We have the ring of characteristic calsses: $\Lambda := \bigoplus \Lambda_m$. The addition structure: if $c_i \in \Lambda_k$, $(c_1+c_2)(B)=c_1(B)+c_2(B)$. the ring structure: if $c_i \in \Lambda_{n_i}$, then we define $(c_1 \cdot c_2)(B) = c_1(B) \cup c_2(B) \in \Lambda_{n_1+n_2}$.
Lemma: For each vector bundle $\xi$ there is a bunlde morphism $(F,g): \xi \rightarrow \gamma_n$ unique upto bundle homotopy, where $\gamma_n=(E,G_n(\Bbb F^\infty), \pi)$ is the Grassmanian bundle.
Now we come to the main statement
The map $H^*(G_n(\Bbb F^\infty), R) \rightarrow \Lambda$, $k$ mapping to the natural transformation it induces is a ring isomorphism.
How does one show this map is injective?
Reference: it is claimed, Theorem 3.3, pg 19 that this is clear. I don't see why.
We see that the map $\Gamma:H^*(G_n(\mathbb{F}^\infty);R)\rightarrow \Lambda$ is injective by evaluating it on the universal example, namely $G_n(\mathbb{F}^\infty)$ itself. Thus if $k\in H^r(G_n(\mathbb{F}^\infty);R)$, then $\Gamma(k)$ is a natural transformation $Vect_n(-)\Rightarrow H^r(-;R)$, and the component we are interested in the the map
$$\Gamma(k)_{G_n(\mathbb{F}^\infty)}:Vect_n(G_n(\mathbb{F}^\infty))\rightarrow H^r(G_n(\mathbb{F}^\infty);R).$$
Then if $\gamma_n$ denotes the universal $n$-plane bundle over $G_n(\mathbb{F}^\infty)$, it is an element of $Vect_n(G_n(\mathbb{F}^\infty))$ which we know to be classified by the identity map $id:G_n(\mathbb{F}^\infty)\rightarrow G_n(\mathbb{F}^\infty)$. Thus from the definition we hav that
$$\Gamma(k)_{G_n(\mathbb{F}^\infty)}[\gamma_n]=id^*k=k$$
is non-zero, and clearly if $h\in H^*(G_n(\mathbb{F}^\infty);R)$ is a second element, then $\Gamma(k)_{G_n(\mathbb{F}^\infty)}[\gamma_n]=\Gamma(h)_{G_n(\mathbb{F}^\infty)}[\gamma_n]$ if and only if $k=h$. It follows from this that the correspondence $\Gamma$ must be injective.