Proving adjoint of exponential is exponential of adjoint

Question

My background is the Spectral Theorem and its corollaries.
Let $A$ a $n \times n$ matrix with complex entries that satisfies $A^* = -A$.
Define $B = e^A$. Show $B^* = e^{-A}$

Since $A$ is normal there exist $U$ unitary such that $$A = U^*DU$$ where $D$ is a diagonal matrix. Then, using $f(x) = e^x$ we have
$$B = f(A) = U^*f(D)U$$ Now, I sense I should be able to prove the result from there, but I am stuck. Any hint would be appreciated. I can't see how $f(A)^* = f(A^*)$

Answer 1

You've almost got it, you have $$B = U^* e^D U$$ so $$B^* = (U^* e^D U)^*= U (e^D)^* U^*.$$ Also, $$(e^D)^*= \left(\sum_{n=0}^\infty \frac{1}{n!} D^n \right)^*$$ which is easily seen to be equal to $$\sum_{n=0}^\infty \frac{1}{n!} (D^*)^n.$$ So $$B^*= U e^{(D^*)} U^*= e^{U D^* U^*} = e^{-A}.$$