Let $X$ be a simplicial complex. The simplicial chain complex $C_*(X)$ is given by: $$ C_k(X) := \{\mathbb{Z}-\textrm{combinations of oriented simplices in X, with the identification }\sigma = - \overline \sigma \}, $$ and the obvious boundary map ($\overline \sigma$ refers to the simplex $\sigma$ with the opposite orientation). The cellular chain is given by assigning to $X$ the obvious CW structure induced by the simplicial structure. We immediately get that we have a natural isomorphism:
$$ H_k(X_k, X_{k-1}) \sim C_k(X) $$
(where $X_i$ is the $i$--skeleton of $X$).
Hence we just need to know if the cellular boundary map: $H_k(X_k, X_{k-1}) \to H_{k-1}(X_{k-1}, X_{k-2})$ given by the "cellular boundary formula" (in hatcher's "algebraic topology" book) corresponds to the boundary map in simplicial homology.
Precise references would be much appreciated (indeed, it is rather curious that I wasn't able to find the answer somewhere for myself).
PS
note that this question refers to the "chain level". Obviously the two cellular and simplicial homologies are isomorphic!
PPS
Having thought a little bit about my question, I suspect that the answer is "obviously yes", and depends on the definition of cellular homology by itself, not on the cellular boundary formula!