Prove that the holonomies along any two homotopic paths are the same if the curvature of the connection vanishes The proof is trivial in the Abelian case by the Stokes' theorem.How to prove it in the non-Abelian case?
 A: A possible answer would be to invoke Ambrose-Singer Holonomy theorem:
Theorem (Ambrose-Singer) Let $M$ be a (smooth) connected manifold, $E\to M$ a vector bundle over $M$, and $\nabla$ a connection on $E$. Then, for each $x\in M$, $\mathfrak{hol}_x(\nabla)$ is a Lie subalgebra of $\text{End}(E_x)$ which, as a vector space, is spanned by all elements of $\text{End}(E_x)$ of 
the form $P_{\gamma}^{-1}[ (F_{\nabla})_y\cdot{(v\wedge w)}] P_{\gamma}$, where $\gamma:[0,1]\to M$ is a piece-wise smooth path with end points $\gamma(0) = x$ and $\gamma(1) = y$, $P_{\gamma}: E_x \to E_y$ is the associated parallel translation map, and $v,w\in T_yM$.
In particular, if the connection $\nabla$ is flat (i.e. if $F_{\nabla}=0$), then $\mathfrak{hol}_x(\nabla)=0$, and therefore the restricted holonomy group $\text{Hol}_x^0(\nabla)$ is trivial, for each $x\in M$. Recalling
$$
\text{Hol}_x^0(\nabla)=\{P_{\gamma}: \gamma\text{ is a null-homotopic loop based at }x\},
$$ the conclusion you need follows. Indeed, given two homotopic paths $\gamma_1$ and $\gamma_2$ (I'm assuming this homotopy preserves the base-points), say with $\gamma_1(0)=\gamma_2(0)=x$, then you can consider the null-homotopic concatenation $\gamma_2^{-1}\cdot{\gamma_1}$, which is a (piece-wise smooth) loop based at $x$. By the above, we have that $1=P_{\gamma_2^{-1}\cdot{\gamma_1}} = P_{\gamma_2}^{-1}\circ P_{\gamma_1}$, i.e. $P_{\gamma_1}=P_{\gamma_2}$, as we wanted.
A principal bundle version of this theorem (together with a proof) can be found e.g. in Kobayashi&Nomizu's book Foundations of differential geometry, vol. I, Theorem 8.1, p. 89. The statement I give here was based on Theorem 2.4.3. (a) of Joyce's book Riemannian Holonomy Groups and Calibrated Geometry.
