Hermitian matrices are unitarily diagonalizable. Suppose we have $A = UDU^{\dagger}$ then $\exp(A) = \sum_{n=1}^\infty\frac{1}{n!}UD^nU^\dagger=Ue^DU^\dagger$ Clearly we have $e^D = \rm{diag}(e^{\lambda_1},\ e^{\lambda_2},\ \cdots,\ e^{\lambda_n})$ Reversing the above process will give us the desired result. Pick $\lambda_i$ so that each $e^{\lambda_i}$ is our desired eigenvalue (this is always positive since we assumed our Hermitian matrix to be positive - definite), we get that any Hermitian matrix is expressible as $e^A$ for some Hermitian $A$.