This question is based on an exercise in Etingof's Introduction to Representation Theory. Let $A$ be an associative algebra (say over a field $k$) and $M$ an $A$-module. Write $\rho : A \to \text{End}_k(M)$ for the action map. By a formal deformation of $M$ I mean a formal power series $\widetilde{\rho} = \rho + \rho_1t + \rho_2t^2 + \cdots$, where each $\rho_i : A \to \text{End}_k(M)$ is a $k$-linear map, satisfying $\widetilde{\rho}(ab) = \widetilde{\rho}(a) \widetilde{\rho}(b)$ for all $a,b \in A$. Two formal deformations are equivalent if they are conjugate by a power series $b = 1 + b_1t + b_2t^2 + \cdots$ where $b_i \in \text{End}_k(M)$.
My question: why does $\text{Ext}^1(M,M) = 0$ imply that any formal deformation $\widetilde{\rho}$ is equivalent to $\rho$? I can see that $\rho_1$ is a $1$-cocycle, so if we are trying to achieve $b\widetilde{\rho} = \rho b$ we should choose $b_1 \in \text{End}_k(M)$ so that $db_1 = \rho_1$. But what happens at the higher levels?