Show that any affine connection $\nabla$ on $\mathbb{R}^n$ is of the form $\nabla=D+\Gamma$, where $D$ is the Euclidean connection and $\Gamma:\mathcal{X}(\mathbb{R}^n) \times \mathcal{X}(\mathbb{R}^n) \rightarrow \mathcal{X}(\mathbb{R}^n)$ is any $C^{\infty}(\mathbb{R}^n)$-bilinear map.
Showing that $D+\Gamma$ is an affine connection is easy, but I don't know how to show the converse. I mean, what's special about $\mathbb{R}^n$?
This follows from the general result that the difference of two affine connections provide such a bilinear map (tensor) $\Gamma$. The nontrivial part is to explain why it is $C^\infty$-linear in the second argument (which is not the case for the connections themselves). But by the Leibniz rule, for any $C^\infty$ function $f$ and vector fields $X,Y$ we have
$$\begin{align}(\nabla - D)(X,fY) &= \nabla(X,fY) - D(X,fY) \\&= (Xf)Y + f\nabla(X,Y) - (Xf)Y - fD(X,Y) \\ &= f(\nabla - D)(X,Y),\end{align}$$
so the $C^\infty$-bilinearity is a consequence of the fact that the "constant terms" (those independent of the connection) cancel each other out in the difference.