= Solution
If $A^\dagger=-A$, then $H=-iA$ is a <Hermitian matrix>, because $H^\dagger=iA^\dagger=-iA=H$. By the <finite-dimensional spectral theorem>, there is a <unitary matrix> $Q$ with
$$
H=Q\operatorname{diag}(\theta_1,\ldots,\theta_n)Q^\dagger,\qquad \theta_j\in\mathbb R.
$$
Thus $A=Q\operatorname{diag}(i\theta_j)Q^\dagger$. Since the <matrix exponential respects similarity>,
$$
\boxed{e^A=QCQ^\dagger,\qquad C=\operatorname{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})}.
$$
Each diagonal entry has modulus one, so $C^\dagger C=I$. Using $Q^\dagger Q=QQ^\dagger=I$,
$$
(e^A)^\dagger e^A=QC^\dagger Q^\dagger QCQ^\dagger=I.
$$
Hence \b[the <exponential of a skew-Hermitian matrix> is unitary].
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