Show that if $\gamma : J \to M$ is a smooth curve in a smooth manifold $M$ of dimension $n$, then $\gamma$ is a smooth immersion if and only if $\gamma'(t) \neq 0$ for all $t \in J$.
Proof: $(\implies)$ We'll prove the forward direction via the contrapositive. Suppose that $\gamma'(t_0) = 0$ for some $t_0 \in J$. Then by definition of $\gamma'$ we have $$\gamma'(t_0) = d\gamma_{t_o}\left(\frac{d}{dt}\bigg|_{t_o}\right) = 0 \in T_{\gamma(t_o)}M.$$ Since $d\gamma_{t_o}$ is a linear map from $T_{t_0} \mathbb{R}$ to $T_{\gamma(t_0)} M$ we have $d\gamma_{t_o}\left(\frac{d}{dt}\bigg|_{t_o}\right) = 0 = d\gamma_{t_0}(0_{T_{t_0}\mathbb{R}})$ thus $d\gamma_{t_0}$ is not injective and so $\gamma$ is not a smooth immersion.
$(\impliedby)$ If $\gamma'(t_0) \neq 0$ for all $t_0 \in J$ then $d\gamma_{t_o}\left(\frac{d}{dt}\bigg|_{t_o}\right) \neq 0$ for all $t_0 \in J$ hence $\operatorname{Im}(d\gamma_{t_0})$ is a $1$-dimensional subspace of $T_{\gamma(t_0)}M$ which has dimension $n$ and since $$\operatorname{rank} d\gamma_{\gamma(t_0)} = \operatorname{dim}(\operatorname{Im}(d\gamma_{t_0})) = 1 = \operatorname{dim}(T_{t_0}\mathbb{R})$$ we have $d_{\gamma_{t_0}}$ to be injective and so $\gamma$ is a smooth immersion. $\square$
Is my proof above correct? If not where have I gone wrong?