Geodesics in $\mathbb{R}^n$ with the trivial connection

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Define geodesic as follows:

Given a tangent bundle $TM\rightarrow M$ with connection $\nabla$, a geodesic is a curve $\gamma:I\rightarrow M$ such that $(\gamma^*\nabla)\dot \gamma = 0$.

(Notice that $(\gamma^*\nabla)\dot \gamma \in \Gamma(\gamma^*TM)$)

I want to check that according to this definition the geodesics in $\mathbb R^n$ with the trivial connection $\nabla(f_1\frac{\partial}{\partial x_1}+...+f_n\frac{\partial}{\partial x_n})=(df_1\frac{\partial}{\partial x_1}+...+df_n\frac{\partial}{\partial x_n}) $ are straight lines $x+tv$.
Take a path $\gamma(t)=(x_1(t),...,x_n(t))$. We have $$ (\gamma^*\nabla)\dot \gamma = (\gamma^*\nabla)( \dot x_1\frac{\partial}{\partial x_1}\vert_{\gamma(-)} +...+\dot x_n\frac{\partial}{\partial x_n}\vert_{\gamma(-)} ) $$ But $$\dot x_i\frac{\partial}{\partial x_i}\vert_{\gamma(-)} = \dot x_i\gamma^*\frac{\partial}{\partial x_i}\vert_{(-)} = \gamma^*(\dot x_i\frac{\partial}{\partial x_i}\vert_{(-)}) $$ So that, using the definition of pullback connection: $$ (\gamma^*\nabla)( \sum_i \gamma^*(\dot x_i\frac{\partial}{\partial x_i}\vert_{(-)})) = \gamma^*(\sum_i \nabla(\dot x_i\frac{\partial}{\partial x_i})) = \gamma^*(\sum_i \frac{d\dot x_i}{dt}dt\frac{\partial}{\partial x_i}) $$ Now I would like to say that this is zero if and only if the $\frac{d\dot x_i}{dt}$ are zero. If I didn't have the $\gamma^*$ that would be clear. But I have no reason to assume that $\gamma^*$ is injective. If it were, I would be done.