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My favorite theorem, the Atiyah-Singer index theorem, seems to have the desired property. The theorem states that the Fredholm index of the Dirac operator on a compact spin manifold $M$ is equal to the $\hat{A}$ genus. There are two essentially different types of proofs: a global, conceptual argument based on little or no calculation; and a detailed local proof involving working with explicit solutions to PDE's. There are many variations and elaborations on the two approaches; here is a basic overview.
Global Proof: One considers the notion of an index map $K(T^*M) \to \mathbb{Z}$ from the K-theory of the tangent bundle of $M$ to the integers, uniquely characterized by a few key axioms. One then constructs two maps which satisfy the axioms (and hence are equal): an analytic index map built using functional analysis and a topological index map built out of an embedding of $M$ into $\mathbb{R}^n$ and the Thom isomorphism in K-theory. The symbol of an elliptic operator $D$ gives rise to an element of $K(T^*M)$; its analytic index is simply the Fredholm index of $D$, while in the case where $D$ is the Dirac operator the topological index can be identified with the $\hat{A}$ genus (upon taking Chern characters).
Local Proof: One first proves that the Fredholm index of $D$ is given by $Tr_s(e^{-t D^2})$, the supertrace of the solution operator for the heat equation for $D$. A standard iterative method for solving the heat equation yields an asymptotic expansion for the smoothing kernel $k_t$ of the heat operator, so since the Fredholm index is independent of $t$ one is lead to try to calculate the constant term in the asymptotic expansion of $tr_s(k_t)$. The strategy (as simplified by Getzler) is to develop a symbol calculus for $D$ which rescales away everything but the constant term. One shows that the appropriate symbol satisfies a certain explicit differential equation (the "quantum mechanical harmonic oscillator") which one can explicitly solve (Mehler's formula). The $\hat{A}$ class manifests itself, as if by magic.