Let $L$ be the Lazard's universal ring, and $R=\mathbb{Z}[b\_1,b\_2,\cdots,b\_n,\cdots]$, regarded as a graded ring with the degree of $b\_i$ equal to $2i$. Let $\theta: L\rightarrow R$ be the homomorphism carrying the universal formal group law $\mu^L$ to the formal group law
$$\mu^R(x\_1,x\_2)=\exp(\log(x\_1)+\log(x\_2)),$$
where the power series 
$$\exp(x)=x+\sum\_{i\geq 1}b\_ix^{i+1},$$
and $\log(x)$ its inverse, denoted as
$$\log(x)=x+\sum\_{i\geq 1}m\_ix^{i+1}.$$
Let $MU$ be the complex cobordism spectrum, and by Quillen's theorem we have the following commutative diagram
$\require{AMScd}$
\begin{CD}
L @>\theta>> R\\
@V \cong V V @VV \cong V\\
\pi\_\*(MU) @>>h> H\_\*(MU;\mathbb{Z})
\end{CD}


where $h$ is the Hurewicz homomorphism. 


In Section 9, Part II of 


*Adams, J. F.*, Stable homotopy and generalised homology, Chicago Lectures in Mathematics. Chicago - London: The University of Chicago Press. X, 373 p. 3.00 (1974). [ZBL0309.55016](https://zbmath.org/?q=an:0309.55016).\*\*


it is stated that the class $[\mathbb{C} P^n]\in\pi\_\*(MU)$ is sent to $(n+1)m\_n\in H\_\*(MU;\mathbb{Z})$ by $h$, and it is indicated there that the argument is a Chern number computation, but I am not seeing the argument.\*\*


I would greatly appreciate your help if you could sketch the proof or point out a reference containing a proof. Thank you!



Complex cobordism and Chern numbers
It is a key result that the composite
$$ MU\_\* \xrightarrow{h} H\_\*(MU;\mathbb Z) \xrightarrow[\sim]{\Phi^{\vee}}H\_\*(BU;\mathbb Z),$$


where $\Phi^{\vee}$ is the dual of the Thom isomorphism $\Phi$, agrees with evaluating on normal Chern numbers. 


In other words, $\langle \Phi(c), h([M])\rangle = \bar c(M)$ for all $c \in H^\*(BU)$ and for all $[M] \in MU\_\*$. 


(A reference in the real case is the diagram on page 228 of *A concise course in algebraic topology* by J.P.May. The complex case is identical.)


So the assertion is just a Chern number calculation: 


$h([\mathbb CP^n]) = (n+1)m\_n$ if and only if, for all $c \in H^{2n}(BU;\mathbb Z)$, $$\langle \Phi(c), (n+1)m\_n\rangle = \bar c(\mathbb CP^n).$$


