(Krull) dimension of any associated graded ring of a ring R equals the dimension of R

Though I heartily agree with Victor Protsak's comment, I will add some references. These might be useful for you, at least if you haven't seen them before. The references add a restriction, however, by assuming that $I$ is an ideal of finite co-length.

Then Corollary 12.5 of Eisenbud's Commutative Algebra uses the theory of Hilbert-Samuel polynomials to prove that $\text{dim}(R)=\text{dim}\text{ gr}_I(R)$.

Alternately, you might also be interested in Corollary 10.12 of the same book. This second corollary assumes that $I=\mathfrak m$, but the proof makes use of "Going down for flat extensions", which has a somewhat different flavor than the Rees ring approach.

This is dealt with in the generality of non-commutative rings in [McConnell, J. C.; Robson, J. C. Noncommutative Noetherian rings. With the cooperation of L. W. Small. Revised edition. Graduate Studies in Mathematics, 30. American Mathematical Society, Providence, RI, 2001. xx+636 pp. MR1811901]