= Solution
For a finite <matrix>, the series defining the <matrix exponential> converges absolutely in any <matrix norm>, because $\|A^m/m!\|\leq\|A\|^m/m!$. Thus the <Hermitian adjoint> may be taken term by term. Since $A^\dagger=A$,
$$
(e^A)^\dagger=\sum_{m=0}^\infty\frac{(A^m)^\dagger}{m!}
=\sum_{m=0}^\infty\frac{(A^\dagger)^m}{m!}
=\sum_{m=0}^\infty\frac{A^m}{m!}=e^A.
$$
Therefore the <matrix exponential> of a <Hermitian matrix> is Hermitian.
If $A=\operatorname{diag}(a_1,\ldots,a_n)$ is a <diagonal matrix>, then $A^m=\operatorname{diag}(a_1^m,\ldots,a_n^m)$. Summing each entry separately gives
$$
\boxed{e^A=\operatorname{diag}(e^{a_1},\ldots,e^{a_n}),\qquad(e^A)_{ij}=\delta_{ij}e^{a_i}.}
$$
For the <Hermitian matrix> case each $a_i$ is real. The diagonal entry formula itself remains valid for complex diagonal entries.
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