Question: For a square matrix $A$. Define $exp(A)=I+\sum\_{n}A^{n}/(n!)$
. I need to prove two things

1. exp(A) converges and is invertible.
2. Its inverse is given by exp(-A).

Second part is straightforward. Can anyone help on first part?

Answer: 
We can show convergence by normal convergence.

$\sum\_{k=0}^{\infty} ||\frac{A^k}{k!}|| \le \sum\_{k=0}^{\infty} \frac{||A||^k}{k!} = e^{||A||}$. Where $||\cdot||$ is a sub-multiplicative norm (e.g. operator norm)

For invertibility, there's probably a nicer way to show this, but I would just show that if $A$ and $B$ commute, then $e^{A+B} = e^{A}e^{B}$.

$e^{A+B} = \sum\_{k=0}^{\infty} \frac{(A+B)^k}{k!} = \sum\_{k=0}^{\infty} \sum\_{i=0}^k \frac{A^i B^{k-i}}{k!} \binom{k}{i} = \sum\_{k=0}^{\infty} \sum\_{i=0}^k \frac{A^i B^{k-i}}{i! (k-i)!} = \sum\_{k=0}^{\infty} \sum\_{i+h=k} \frac{A^i B^{h}}{i! h!} = \sum\_{i=0}^{\infty} \sum\_{h=0}^{\infty} \frac{A^i B^{h}}{i! h!} = e^A e^B$

$A$ and $-A$ commute so $e^A e^{-A}= e^{A - A} = e^{0} = I$
It's possible to do this rather cutely with [Jacobi's formula](https://en.wikipedia.org/wiki/Jacobi%27s_formula), which can be used to show that: $$ \det(e^{cA})=e^{\text{tr}(cA)} $$

Let $c=1$. Then $\det(e^{A})=e^{\text{tr}(A)}$, but $e^{\text{tr}(A)}>0$, so $\det(e^A)\ne 0$, which means $e^A$ must be invertible. (For convergence, I don't think one can do better than using the sub-multiplicativity of an operator norm, as the other answer does).