= Solution
Let $H$ be a <Hermitian matrix> and $He=\lambda e$ with $e\ne0$. The <inner product> $e^\dagger He$ is real because its conjugate equals $e^\dagger H^\dagger e=e^\dagger He$. But it also equals $\lambda e^\dagger e$, with $e^\dagger e>0$. Thus every <eigenvalue> is real. For two <eigenvectors>,
$$
e_i^\dagger He_j=\lambda_j e_i^\dagger e_j
=(He_i)^\dagger e_j=\lambda_i e_i^\dagger e_j.
$$
When $\lambda_i\ne\lambda_j$, this proves $e_i^\dagger e_j=0$, so the <eigenvectors> are orthogonal.
For a real <antisymmetric matrix>, $(iA)^\dagger=-iA^T=iA$, so $iA$ is a <Hermitian matrix>. Its real <eigenvalues> imply that the <eigenvalues> of $A$ are purely imaginary. To ensure a nonzero one without assuming it, write
$$
A=\begin{pmatrix}0&-w_3&w_2\\w_3&0&-w_1\\-w_2&w_1&0\end{pmatrix},\qquad w\ne0.
$$
Expansion of the <characteristic polynomial> gives $\det(tI-A)=t(t^2+|w|^2)$. Therefore the nonzero <eigenvalue> $-i\theta$ exists with $\theta=|w|>0$, and a nonzero complex <eigenvector> $e_1=u+iv$ satisfies
$$
A(u+iv)=-i\theta(u+iv)=\theta v-i\theta u.
$$
Separating real and imaginary parts gives
$$
\boxed{Au=\theta v,\qquad Av=-\theta u.}
$$
Neither real vector can vanish, since either vanishing would force the other to vanish. Antisymmetry yields $0=u\cdot Au=\theta u\cdot v$ and
$$
-\theta|u|^2=u\cdot Av=-(Au)\cdot v=-\theta|v|^2.
$$
Thus $u,v$ are orthogonal and have equal length.
In odd dimension $\det A=\det A^T=\det(-A)=-\det A$, so $\det A=0$. Its real homogeneous system therefore has a nonzero real <eigenvector> $e_3$ with $Ae_3=0$. It is perpendicular to $u,v$, because $e_3\cdot Au=-(Ae_3)\cdot u=0$ and similarly for $Av$. Normalize it and choose its sign so $(u/|u|,v/|v|,e_3)$ is positively oriented; this is an <orthonormal basis>.
The <matrix exponential> series converges absolutely, allowing even and odd terms to be grouped. Since $A^2u=-\theta^2u$ and $A^2v=-\theta^2v$, it acts as
$$
Ru=u\cos\theta+v\sin\theta,\qquad
Rv=v\cos\theta-u\sin\theta,\qquad Re_3=e_3.
$$
Consequently the <skew-symmetric exponential as an axial rotation> has, in the displayed <orthonormal basis>, the matrix
$$
\boxed{\begin{pmatrix}\cos\theta&-\sin\theta&0\\\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix}.}
$$
It is orthogonal with determinant one, fixes its axis and rotates the perpendicular plane. Hence \b[$R=e^A$ is a rotation matrix], with angle $\theta$ about the chosen oriented axis.
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