Matrix exponential commutes with conjugation

Statement
Suppose $$K$$ is a field and $$A,B$$ are $$n \times n$$ matrices over $$K$$ such that the matrix exponential of $$A$$, denoted $$\exp (A)$$, exists. Then, the matrix exponential of $$BAB^{-1}$$ also exists, and:

$$\! \exp(BAB^{-1}) = B\exp(A)B^{-1}$$

Related facts

 * Exponential map commutes with adjoint action is a version for Lie groups and Lie algebras.
 * Exponential of adjoint map corresponds to conjugation by group element for Lazard Lie ring is the version for the Lazard correspondence.