= Solution
The <matrix exponential> power series is absolutely convergent for a finite matrix, so taking the <conjugate transpose> term by term is valid. Since $M^\dagger=M$,
$$
U^\dagger=\left(\sum_{n=0}^\infty\frac{i^nM^n}{n!}\right)^\dagger
=\sum_{n=0}^\infty\frac{(-i)^nM^n}{n!}=e^{-iM}.
$$
The absolute convergence also justifies the <Cauchy product>. In $e^{-iM}e^{iM}$, the coefficient of $M^n$ is proportional to $\sum_{k=0}^n(-1)^k\binom nk$, which is zero for $n>0$ and one for $n=0$, by the <binomial theorem>. Consequently $U^\dagger U=I$; the reversed product is also $I$. This proves that the exponential is <unitary> directly from the power series, without invoking diagonalization.
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