Proof that the Hodge-de Rham Rank Equals the Euler Characteristic

Can someone please provide a good (online accessible) reference for the well-known identity $$\text{rank((d + d}^*)^+) = \sum_{i=}^n (-1)^i \dim(H^i(M)),$$ where $M$ is a manifold of dimension $n$, with d and d$^*$ its exterior derivative and co-derivative respectively, the $+$-superscript denoting restriction of the operator to the positive forms, and finally $H^i(M)$ the $i$-th cohomology group. (If someone would like to post the proof as an answer I wouldn't object, but I guess it's too well-known for that.)

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I'm not entirely sure what exactly your left side means, but I'll try for the closest fit. If $M$ is compact oriented and Riemannian, the Hodge theorem gives that the space of $C^\infty$ $i$-forms decomposes as an orthogonal direct sum $(\text{harmonic forms})\oplus im(d)\oplus im(d^\ast)$. This will imply that if you consider the $d+d^\ast$ as a map from the space of even degree forms to odd degree forms, then the index is exactly the Euler characteristic on the right side of your equation. I don't know about an online ref. but lots of books (Griffiths-Harris, de Rham, Warner, Wells...) give a proof of the Hodge theorem.
Added... In more detail $$index(d+d^\ast) = \dim \ker (d+d^\ast)|_{\text{even forms}}-\dim im (d+d^\ast)^\perp |_{\text{odd forms}}$$ The above kernel is the space of even harmonic forms, which can be identified with $H^{even}(M)$ by the Hodge theorem. The second space can be identified with $H^{odd}(M)$ for similar reasons.
P.S. By index of an operator $A$ I mean dim of the image of $A$ minus the dim of the cokernel of $A$. – user33845 May 9 '13 at 14:17