[proofplan] The proof is a direct comparison of characteristic class evaluations. The complex bundle isomorphism $TM\cong f^*TN$ identifies the Chern classes of $TM$ with the pullbacks of the Chern classes of $TN$, by [naturality of Chern classes](/theorems/9771). Multiplicativity of pullback then identifies every Chern monomial on $M$ with the pullback of the corresponding Chern monomial on $N$. Finally, because $f$ is orientation-preserving, it sends the fundamental class of $M$ to the fundamental class of $N$, so the top-degree pairings agree. [/proofplan]

[step:Fix an arbitrary Chern monomial of top degree] Let $(a_1,\dots,a_n)$ be a finite sequence of non-negative integers satisfying \begin{align*} \sum_{i=1}^n i a_i=n. \end{align*} Define the cohomology class \begin{align*} \alpha_M:=c_1(TM)^{a_1}\cdots c_n(TM)^{a_n}\in H^{2n}(M;\mathbb Z) \end{align*} and define the corresponding cohomology class \begin{align*} \alpha_N:=c_1(TN)^{a_1}\cdots c_n(TN)^{a_n}\in H^{2n}(N;\mathbb Z). \end{align*} The degree condition ensures that each product lies in top real cohomological degree $2n$, since $c_i$ has degree $2i$. [/step]

[step:Use the complex bundle isomorphism to identify Chern classes]For each $i\in\{1,\dots,n\}$, the isomorphism $\Phi:TM\to f^*TN$ is an isomorphism of complex vector bundles over $M$. Therefore isomorphic complex vector bundles have the same Chern classes, and [Naturality of Chern Classes][citetheorem:9771] gives \begin{align*} c_i(TM)=c_i(f^*TN)=f^*c_i(TN) \end{align*} in $H^{2i}(M;\mathbb Z)$.[/step]

[guided]We need to convert the bundle hypothesis into a statement about cohomology classes. The hypothesis provides a complex vector bundle isomorphism \begin{align*} \Phi:TM\longrightarrow f^*TN \end{align*} over $M$. Since Chern classes are invariants of complex vector bundle isomorphism, this gives \begin{align*} c_i(TM)=c_i(f^*TN) \end{align*} for every $i\in\{1,\dots,n\}$. Now apply [Naturality of Chern Classes][citetheorem:9771] to the smooth map $f:M\to N$ and the complex vector bundle $TN\to N$. Its hypotheses are satisfied: $f$ is smooth because it is a diffeomorphism, and $TN\to N$ is a finite-rank complex vector bundle because $N$ is a complex manifold. Hence \begin{align*} c_i(f^*TN)=f^*c_i(TN) \end{align*} for every $i\in\{1,\dots,n\}$. Combining the two equalities gives \begin{align*} c_i(TM)=f^*c_i(TN) \end{align*} in $H^{2i}(M;\mathbb Z)$.[/guided]

[step:Pull back the entire Chern monomial] The pullback \begin{align*} f^*:H^*(N;\mathbb Z)\longrightarrow H^*(M;\mathbb Z) \end{align*} is a graded ring homomorphism. Therefore, using the identities from the previous step, \begin{align*} \alpha_M = \prod_{i=1}^n c_i(TM)^{a_i} = \prod_{i=1}^n \bigl(f^*c_i(TN)\bigr)^{a_i} = f^*\left(\prod_{i=1}^n c_i(TN)^{a_i}\right) = f^*\alpha_N. \end{align*} Thus the Chern monomial on $M$ is the pullback of the corresponding Chern monomial on $N$. [/step]

[step:Transfer the top-degree pairing through the orientation-preserving diffeomorphism] Since $f:M\to N$ is orientation-preserving with respect to the complex orientations, its induced map on singular homology satisfies \begin{align*} f_*[M]=[N]\in H_{2n}(N;\mathbb Z). \end{align*} By naturality of the cohomology-homology pairing, applied to the class $\alpha_N\in H^{2n}(N;\mathbb Z)$ and the fundamental class $[M]\in H_{2n}(M;\mathbb Z)$, \begin{align*} \left\langle f^*\alpha_N,[M]\right\rangle = \left\langle \alpha_N,f_*[M]\right\rangle. \end{align*} Using $f_*[M]=[N]$ and $\alpha_M=f^*\alpha_N$, we obtain \begin{align*} \left\langle \alpha_M,[M]\right\rangle = \left\langle f^*\alpha_N,[M]\right\rangle = \left\langle \alpha_N,[N]\right\rangle. \end{align*} By the definitions of $\alpha_M$ and $\alpha_N$, this is exactly \begin{align*} \left\langle c_1(TM)^{a_1}\cdots c_n(TM)^{a_n},[M]\right\rangle = \left\langle c_1(TN)^{a_1}\cdots c_n(TN)^{a_n},[N]\right\rangle. \end{align*} Because the sequence $(a_1,\dots,a_n)$ was arbitrary subject to $\sum_{i=1}^n i a_i=n$, all Chern numbers of $M$ and $N$ agree. [/step]