 Solution

The matrix exponential power series is absolutely convergent for a finite matrix, so taking the conjugate transpose term by term is valid. Since $M^{†}=M$,The absolute convergence also justifies the Cauchy product. In $e^{-iM}{e}^{iM}$, the coefficient of $M^n$ is proportional to ${\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})$, which is zero for $n>0$ and one for $n=0$, by the binomial theorem. Consequently $U^{†}U=I$; the reversed product is also $I$. This proves that the exponential is unitary directly from the power series, without invoking diagonalization.