So I'm confused about a part of the proof Lee gives for zero curvature implies flatness, this is theorem 7.3.
Proof:Let ${E_1,...,E_n}$ be an orthonormal basis at $p\in M$. Let $(x^i)$ be a coordinate system where $E_i\mid_p=\frac{\partial}{\partial x^i}$. By shrinking the coordinates neighborhood. We may assume that the image of the coordinates chart is a cube $C_\epsilon=\{x:|x^i|<\epsilon,i=1,...,n\}$. Begin by parallel translating on the $x^1$-axis. Then proceed to parallel translate along the $x^2$-axis. Continuing this process we get n vector fields on $C_\epsilon$.
Now we want to show that the frame is parallel. We want to show that
$$\nabla_{\partial_1}E_j=...=\nabla_{\partial_k}E_j\equiv 0$$ on $M_k=\{x^{k+1}=...=x^n=0\}$.By induction assume that it holds for some $k$. On $M_{k+1}$, $\nabla_{\partial_{k+1}}E_j=0$ by construction, and for $i\leq k$, $\nabla_{\partial_i}E_j=0$ on the hyperplane where $x^{k+1}=0$ by the inductive hypothesis. So it suffices to show that $\nabla_{\partial_{k+1}}(\nabla_{\partial_i}E_j)\equiv 0$.
Question: So how is the last equation,$\nabla_{\partial_{k+1}}(\nabla_{\partial_i}E_j)\equiv 0$ sufficient to showing the condition given above?
The inductive hypothesis implies that $\nabla_{\partial_i} E_j\equiv 0$ on $M_k$ for $1\le i \le k$. So if we can show that $\nabla_{\partial_i} E_j$ is parallel along the $x^{k+1}$-curves starting on $M_k$, then it is equal to the parallel transport of the zero vector and thus is zero on each $x^{k+1}$-curve. Since every point of $M_{k+1}$ is on one of these curves, it follows that $\nabla_{\partial_{i}}E_j=0$ on all of $M_{k+1}$.
(I realize that my explanation of this point was a bit cryptic. In the second edition of the book, which hopefully will appear sometime in 2017, I'll try to explain it a bit more fully.)