It is well known that for every unitary operator $\hat U$ an exponential of the form $$ \hat U = e^{i\hat H} $$
exists ($\hat H$ is hermitian). But I can only prove it the other way round:
$$ (e^{i\hat A})^\dagger = \sum_{n=0}^\infty \frac{(-i)^n(A^n)^\dagger}{n!}=e^{-i\hat A^\dagger}=e^{-i\hat A} $$ with $A$ hermitian.
Now suppose $$ \hat U\hat U^\dagger = e^{i\hat A} e^{-i\hat A}=1 $$
so $\hat U$ has to be unitary. So now I have proven the statement that for every hermitian operator there exists a unitary operator, right?
Now how do I know that I can always find an exponential form of a unitary operator? Is this statement eventually true "in both directions"?
Use the spectral theorem for normal operators. $H = -i \log(U)$ (for any branch of the logarithm).