Let $G$ be a Lie group. The exponential map
$$\exp: T_e G \to G$$
is defined by $\exp (X)=c_X(1)$, where $c_X$ is the maximal integral curve with initial point $e$ of the left invariant vector field $v_X$ on $G$ determined by $v_X(e)=X$, or more specifically $v_X(e)=d_e(l_x)X,$ where $l_x$ is the left translation by $x$ on the Lie group $G$.
I wonder how to prove by only using the above definitions that when we take the differential of $\exp:T_e G \to G$, we get
$$d_0(\exp)X=\frac{d}{dt}\exp(tX)|_{t=0}. $$
Note $d_0(\exp)$ is a map $T_0(T_e G)\to T_e G$.
This is the definition of the differential. Recall that when $f: M\to N$ is a differentiable mapping and $x\in M$, then $$d_x f : T_xM \to T_{f(x)}N$$ is defined as follows: let $X\in T_xM$. Then there is a smooth curve $\gamma : (-\epsilon, \epsilon) \to M$ so that $\gamma(0) = x$ and $\gamma'(0) = X$ (indeed, this is more or less the definition of a tangent vector). Then by definition $$ d_x f (X) = \frac{d}{dt} (f\circ \gamma)(t) \bigg|_{t=0}.$$
In your example $\exp : T_eG\to G$, then we have $d_0 \exp : T_0(T_e G) \to T_eG$, given by $$ d_0 \exp (X) = \frac{d}{dt} \exp (\gamma (t))\bigg|_{t=0}$$ when one choose $\gamma (t) = tX$, we recover your equation.