The Baker-Campbell-Hausdorff formula says that
$$
e^{tX}e^{tY}=e^Z
$$
where
$$
Z=tX+tY+\frac{t^2}{2}[X,Y]+\frac{t^3}{12}\big([X,[X,Y]]-[Y,[X,Y]]\big)+\dots
$$
and further terms involve higher powers of $t$ and higher order commutators. Since all commutators of order higher than $1$ are $0$, this gives the result. That is, because $[X,Y]$ commutes with both $X$ and $Y$,
$$
\begin{align}
e^{tX}e^{tY}
&=e^{tX+tY+\frac{t^2}{2}[X,Y]}\\
&=e^{tX+tY}e^{\frac{t^2}{2}[X,Y]}
\end{align}
$$
The Baker-Campbell-Hausdorff Formula
Suppose $X$ and $Y$ are elements in a non-commutative algebra. Define the operator $\mathrm{ad}(X)$ by $\mathrm{ad}(X)Y = [X,Y] = XY-YX$. Then we get the following:
Lemma 1: $\displaystyle X^kY=\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^jYX^{k-j}$
Proof: The case $k=0$ is trivial. Suppose the statement is true for some $k$, then
$$
\begin{align}
X^{k+1}Y
&=\mathrm{ad}(X)\left(X^kY\right)+\left(X^kY\right)X\\
&=\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^{j+1}YX^{k-j}
+\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^jYX^{k-j+1}\\
&=\sum_{j=0}^{k+1}\binom{k}{j-1}\mathrm{ad}(X)^{j}YX^{k-j+1}
+\sum_{j=0}^{k+1}\binom{k}{j}\mathrm{ad}(X)^jYX^{k-j+1}\\
&=\sum_{j=0}^{k+1}\binom{k+1}{j}\mathrm{ad}(X)^jYX^{k-j+1}
\end{align}
$$
Thus, the statement is true for $k+1$.$\quad\square$
Let $\mathrm{D}$ be the usual derivative: $\mathrm{D}(XY) = X\,\mathrm{D}(Y) +\mathrm{D}(X)\,Y$. Then
$$
\begin{align}
\mathrm{D}\left(X^n\right)
&=\sum_{k=0}^{n-1}X^k\mathrm{D}(X)X^{n-k-1}\\
&=\sum_{k=0}^{n-1}\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^j\mathrm{D}(X)X^{k-j}X^{n-k-1}\\
&=\sum_{j=0}^{n-1}\sum_{k=j}^{n-1}\binom{k}{j}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\\
&=\sum_{j=0}^{n-1}\binom{n}{j+1}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\tag{1}
\end{align}
$$
Using the power series for $e^X$, we get
$$
\begin{align}
\mathrm{D}\left(e^X\right)
&=\sum_{n=0}^\infty\sum_{j=0}^{n-1}\frac1{n!}\binom{n}{j+1}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\\
&=\sum_{j=0}^\infty\sum_{n=j+1}^\infty\frac1{(j+1)!}\frac1{(n-j-1)!}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\\
&=\sum_{j=0}^\infty\sum_{n=0}^\infty\frac1{(j+1)!}\frac1{n!}\mathrm{ad}(X)^j\mathrm{D}(X)X^n\\
&=\left[\frac{e^{\mathrm{ad}(X)}-1}{\mathrm{ad}(X)}\mathrm{D}(X)\right]e^X\tag{2}
\end{align}
$$
Because $\frac{\mathrm{d}}{\mathrm{d}t}e^{tX}=Xe^{tX}$, we get the following
Lemma 2: $e^{tX}Ye^{-tX}=e^{t\,\mathrm{ad}(X)}Y$
Proof: Note that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{tX}Ye^{-tX}\right)
&=X\left(e^{tX}Ye^{-tX}\right)-\left(e^{tX}Ye^{-tX}\right)X\\
&=\mathrm{ad}(X)\left(e^{tX}Ye^{-tX}\right)
\end{align}
$$
Thus, $e^{tX}Ye^{-tX}=e^{t\,\mathrm{ad}(X)}Y$.$\quad\square$
From this, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{tX}e^{tY}\right)
&=X\left(e^{tX}e^{tY}\right)+\left(e^{tX}Ye^{tY}\right)\\
&=\left(X+e^{t\,\mathrm{ad}(X)}Y\right)\left(e^{tX}e^{tY}\right)\tag{3}
\end{align}
$$
Using the power series for $\log$, we see that formally, there is a $Z$ so that $e^Z=e^{tX}e^{tY}$.
Combining
$$
\frac{\mathrm{d}}{\mathrm{d}t}\left(e^Z\right)e^{-Z}
=\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{tX}e^{tY}\right)e^{-tY}e^{tX}\tag{4}
$$
with $(2)$ and $(3)$, we get that
$$
\frac{e^{\mathrm{ad}(Z)}-1}{\mathrm{ad}(Z)}\frac{\mathrm{d}}{\mathrm{d}t}Z
=X+e^{t\,\mathrm{ad}(X)}Y\tag{5}
$$
and thus,
$$
\frac{\mathrm{d}}{\mathrm{d}t}Z
=\frac{\mathrm{ad}(Z)}{e^{\mathrm{ad}(Z)}-1}\left(X+e^{t\,\mathrm{ad}(X)}Y\right)\tag{6}
$$
If we iterate $(6)$, the power series in $t$ for $Z$ converges at least one order of $t$, hence one order of commutator, per iteration to yield
$$
Z=tX+tY+\frac{t^2}{2}[X,Y]+\frac{t^3}{12}\big([X,[X,Y]]-[Y,[X,Y]]\big)+\dots\tag{7}
$$
where only terms with commutators of third or higher order have been omitted.
