First let $L^{\bullet}$ be a pro-nilpotent differential graded Lie algebra (dgla). We have the set of Maurer-Cartan elements in $L^{\bullet}$ ($MC(L^{\bullet})$) which are $\alpha \in L^1$ such that it satisfies the Maurer-Cartan equation $$ \partial \alpha+ \frac{1}{2}[\alpha,\alpha]=0. $$ We have a definition of gauge equivalance: $\alpha_0,\alpha_1\in MC(L^{\bullet})$ are called gauge equivalent if and only if there exists a $\xi \in L^0$ such that $$ e^{\text{ad}\xi}\circ(\partial +\text{ad}\alpha_0)\circ e^{-\text{ad}\xi}=\partial +\text{ad}\alpha_1 $$ or in other words $$ e^{\text{ad}\xi}\alpha_0-\frac{e^{\text{ad}\xi}-1}{\text{ad}\xi}\partial\xi=\alpha_1. $$ From the first definition it is easy to see that gauge equivalence is really an equivalent relation. From the second definition we can define a path between $\alpha_0$ and $\alpha_1$. Let $$ \alpha(t)=e^{t\text{ad}\xi}\alpha_0-\frac{e^{t\text{ad}\xi}-1}{\text{ad}\xi}\partial\xi. $$ Then $\alpha(t)$ is a power series of $t$ in $L^{\bullet}$, $\alpha(0)=\alpha_0$, $\alpha(1)=\alpha_1$ and we can prove $\partial\alpha(t)+ \frac{1}{2}[\alpha(t),\alpha(t)]=0.$

Now we come to $L_{\infty}$ algebra $L^{\bullet}$ with higher bracket $[\cdot,\ldots,\cdot]_n$ with $n$ auguments. We still have Maurer-Cartan elements in $ L^{\bullet} $ ( $MC(L^{\bullet})$) which are $ \alpha \in L^1$ such that it satisfies the Maurer-Cartan equation $$ \partial \alpha+ \sum \frac{1}{k!}[\alpha,\ldots,\alpha]_k=0. $$

My question is how to define equivalence of Maurer-Cartan elements in $ L^{\bullet} $?

Of course we can define $\alpha_0,\alpha_1\in MC(L^{\bullet})$ are "equivalent" if and only if there exists a power series $\alpha(t)\in L^{\bullet}$ such that $\alpha(0)=\alpha_0$, $\alpha(1)=\alpha_1$ and $\alpha(t)$ satisfies the $L_{\infty}$ Maurer-Cartan equation. However, it is difficult to show that this is an equivalent relation, for example, how to connect two paths?

It seems that a generalization of gauge equivalence is what we want. But $e^{\text{ad}\xi}$ is not enough since we have higher bracket.