TITLE: Prove that $e^{-A} = (e^{A})^{-1}$
QUESTION [6 upvotes]: Let $A, B \in R^{n \times n}$. Prove that $e^{-A} = (e^{A})^{-1}$.
($R$ is the real numbers)
I've tried messing around with both sides, evaluated as sums. I just can't get the two to match up.
Any ideas?

REPLY [5 votes]: Another less elegant way:
$$e^A = I+ A + \frac{A^2}{2!}+ \frac{A^3}{3!}+\cdots$$
$$e^{-A} = I - A + \frac{A^2}{2!}-\frac{A^3}{3!}+\cdots$$
$$e^A \, e^{-A} = I + \sum_{k=1}^\infty b_k A^k$$
where $$b_k =\sum_{j=0}^k \frac{(-1)^j}{j! (k-j)!}= \frac{1}{k!}\sum_{j=0}^k (-1)^j {k \choose j}=\frac{1}{k!}(1-1)^k =0$$