Given a matrix Lie group $G$, one way to define the associated Lie algebra is as $$\operatorname{Lie}(G)=\{ X\in M(n;\mathbb C) | \,e^{tX}\in G\, \text{ for all }\, t\in\mathbb R\}.\tag1$$ Now, we also know that matrix Lie algebras are closed under commutator. So this means that $X,Y\in \operatorname{Lie}(G)$ implies $[X,Y]=XY-YX\in\operatorname{Lie}(G)$. Using the above definition of Lie algebra, this means that $e^{tX},e^{tY}\in G$ for all $t$ implies that also $e^{t[X,Y]}\in G$ for all $t$.
I've usually seen this proven in specific examples of matrix Lie groups. But how can it be proven in full generality? I'm guessing some application of the BCH formula, but I don't quite see how to use it here.
Usually, you don't directly prove that $e^{t[X,Y]}\in G$ for all $t$, but you just show that the Lie algebra of $G$ is closed under the commutator, which then implies the claim by (1). To do this, you observe that for $X\in \operatorname{Lie}(G)$ and $A\in G$, you get $e^{tAXA^{-1}}=Ae^{tX}A^{-1}$, so $AXA^{-1}\in \operatorname{Lie}(G)$.
Knowing this, you see that for $X,Y\in \operatorname{Lie}(G)$, $c(t):=e^{tX}Ye^{-tX}$ is a smooth curve in $\operatorname{Lie}(G)$, which is a linear subspace in $M_n(\mathbb R)$ (which is not obvious from (1) but can be proved). But this implies that also $c'(0)\in \operatorname{Lie}(G)$ and a direct computation shows that $c'(0)=XY-YX$, which implies the claim.