Let $G$ be Lie group with the Lie algebra denoted $\mathfrak{g}$. For $X\in \mathfrak{g}$ calculate the composition $$\mathfrak{g}\cong T_{X}\mathfrak{g} \overset{\underrightarrow{(d\exp)X}}{} T_{\exp(X)}G \overset{\underrightarrow{dL_{\exp(X)^{-1}}}}{} T_{e}G=\mathfrak{g}.$$ Namely, show that it is equal to $$Y \mapsto \frac{1+\exp(\mbox{ad}(X))}{\mbox{ad}(X)}(Y)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)!}(\mbox{ad}(X)^{n})(Y)$$ (Where the powerseries $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)!}(\mbox{ad}(X)^{n})$ converges absolutely and uniformly on compact sets, thus defining an analytic function $\mathfrak{g}\rightarrow \mbox{End}(\mathfrak{g})$. )
Remark: Let $\Phi:\mathfrak{g} \rightarrow \mathfrak{g}$ the map in question. If $c:(-\epsilon,\epsilon)\rightarrow \mathfrak{g}$ is a curve with $c(0)=X$ and $\dot{c}(0)=Y$, then I prove that $$\Phi(Y)=\left.\frac{d}{dt}\right|_{t=0} \exp(X)^{-1}\exp(c(t))$$ (for example, $c(t)=X+tY$).
Consider $$\Psi(s,t):=\frac{\partial}{\partial t}\exp(sX)^{-1}\exp(s\cdot c(t)).$$ Note that our $\Phi$ is a special value of $\Psi(1,t)$, then we have calculate $\Psi(1,t)$, in a book find $$\Psi(1,t)=\int_{0}^{1}\frac{\partial \Psi}{\partial s}ds.$$ I want to use this last expression, but to arrive at the desired result I must calculate $\frac{\partial \Psi}{\partial s}$ to find something like $\mbox{Ad}(\exp(-sX))(\dot{c}(t))$. The following would be easy, but I have not been able to prove the latter.