at.algebraic topology - singular homology of a differential manifold

Let M is differential manifold, $Delta$ is the closed simplex $[p_0, p_1,...,p_k]$. A differential singular k-simplex $sigma$ of M is a smooth mapping $sigma:Delta -> M$.

And we construct a chain complex in the same way we construct the chain complex of singular homology, we gain its homology group.

My question is why this homology group equals the singular homology group? I have tried finding in many books but there is no answer of this question.