= Solution
Introduce a parameter $t$ to keep track of total degree. The <power series> for the <matrix exponential> gives
$$
e^{tX}e^{tY}=I+t(X+Y)+t^2\left(\frac{X^2}{2}+XY+\frac{Y^2}{2}\right)+O(t^3).
$$
Set $S=X+Y$ and $C=\tfrac12[X,Y]$, where the <commutator> is $[X,Y]=XY-YX$. Expanding a second <matrix exponential> gives
$$
e^{tS+t^2C}=I+tS+t^2\left(C+\frac{S^2}{2}\right)+O(t^3).
$$
Since
$$
C+\frac{S^2}{2}=\frac12(XY-YX)+\frac12(X^2+XY+YX+Y^2)=\frac{X^2}{2}+XY+\frac{Y^2}{2},
$$
the two expressions agree through total degree two. More explicitly, applying the local <matrix logarithm> to the first expansion gives $\log(e^{tX}e^{tY})=t(X+Y)+\tfrac12t^2[X,Y]+O(t^3)$. This proves the displayed order of the <Baker--Campbell--Hausdorff formula>. The <Lie algebra> is closed under the <commutator>, so its displayed exponent lies in the same <Lie algebra>. The expansion is a <formal power series> identity, or a convergent identity sufficiently near the <identity matrix>; this argument does not assert unrestricted global convergence.
The next two terms of the <Baker--Campbell--Hausdorff formula> are the two cubic <commutators>:
$$
\boxed{\log(e^Xe^Y)=X+Y+\frac12[X,Y]+\frac1{12}[X,[X,Y]]+\frac1{12}[Y,[Y,X]]+O(4).}
$$
Here $O(4)$ means total degree at least four in $X,Y$. If one continues by one more degree, the quartic contribution is $-\tfrac1{24}[Y,[X,[X,Y]]]$.
For the <Pauli matrices>, use the <Pauli matrix multiplication law>
$$
\tau_a\tau_b=\delta_{ab}I+i\epsilon_{abc}\tau_c.
$$
Put $N=\mathbf n\cdot\boldsymbol\tau$. The antisymmetric term disappears when contracted with $n_an_b$, so $N^2=|\mathbf n|^2I=I$. Consequently every even power of $N$ equals $I$, and every odd power equals $N$. Splitting the <matrix exponential> into even and odd terms of its <power series> yields
$$
\boxed{e^{i\alpha N}=\cos\alpha\,I+i\sin\alpha\,N.}
$$
The <Pauli matrices> are <Hermitian matrices>, and the components of $\mathbf n$ are real, so $N^\dagger=N$. For real $\alpha$, the resulting matrix $U$ satisfies
$$
U^\dagger U=(\cos\alpha I-i\sin\alpha N)(\cos\alpha I+i\sin\alpha N)=I.
$$
Also $\operatorname{tr}N=0$ and $N^2=I$, so its two <eigenvalues> are $1,-1$. The <eigenvalues> of $U$ are therefore $e^{i\alpha},e^{-i\alpha}$, whose product is one. Thus $U$ is a <unitary matrix> with <determinant> one: \b[it belongs to the <SU(2) group>].
Multiplying the two exact <Pauli matrix> exponentials, and using $\tau_1\tau_2=i\tau_3$, gives
$$
\boxed{e^{i\alpha\tau_1}e^{i\beta\tau_2}=\cos\alpha\cos\beta\,I+i\left(\sin\alpha\cos\beta\,\tau_1+\cos\alpha\sin\beta\,\tau_2-\sin\alpha\sin\beta\,\tau_3\right).}
$$
In particular, the $\tau_3$ term has a minus sign. Its quadratic <Taylor expansion> is
$$
I+i\alpha\tau_1+i\beta\tau_2-i\alpha\beta\tau_3-\frac12(\alpha^2+\beta^2)I+O(3).
$$
Taking $X=i\alpha\tau_1$, $Y=i\beta\tau_2$, the <Pauli matrix commutator identity> gives $[X,Y]=-2i\alpha\beta\tau_3$. The quadratic exponent in the <Baker--Campbell--Hausdorff formula> is therefore $Z=i\alpha\tau_1+i\beta\tau_2-i\alpha\beta\tau_3$. In $e^Z=I+Z+\tfrac12Z^2+O(3)$, only the linear part of $Z$ contributes to $Z^2$ through degree two. The anticommutator $\tau_1\tau_2+\tau_2\tau_1=0$ gives $Z^2=-(\alpha^2+\beta^2)I+O(3)$, reproducing the exact product's quadratic expansion.
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