= Solution
Since $A$ is <Skew-Hermitian>, $H=iA$ is a <Hermitian matrix>. The <spectral theorem> gives a <unitary matrix> $Q$ with $Q^\dagger HQ=\operatorname{diag}(\lambda_1,\ldots,\lambda_N)$ and real <eigenvalues> $\lambda_j$. Hence $Q^\dagger AQ=\operatorname{diag}(-i\lambda_j)$ and
$$
\boxed{|\det(I+A)|^2=\prod_{j=1}^N|1-i\lambda_j|^2=\prod_{j=1}^N(1+\lambda_j^2)\geq1.}
$$
In particular the <determinant> is nonzero, so $I+A$ is invertible. Applying the same argument to $-A$ proves that $I-A$ is also invertible. Equality holds precisely when all $\lambda_j=0$, equivalently $A=0$.
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