Let $A\in M_{n\times n}(\mathbb{C})$. We define
$\displaystyle\ln A=\sum_{k=1}^\infty(-1)^{k+1}\frac{(A-I)^k}{k}$
This series is convergent in particular when $A-I$ is nilpotent. Is it true (and how do you prove it) that if $A-I$ is nilpotent, then $e^{\ln A}=A$? (Also what about $\ln e^A=A$?)
For convenience, let $N = I-A$ and suppose $N^M=0$. Consider $$\ln(I-tN) = - \sum_{k=1}^{M-1} (tN)^k /k$$ Note that $$ \dfrac{d}{dt} \ln(I-tN) = - \sum_{k=1}^{M+1} (tN)^{k-1} = -(I-tN)^{-1}$$ Let $g(t) = \exp(\ln(I-tN))$. Then $g(0) = \exp({\bf 0}) = I$ and $$ g'(t) = g(t) \dfrac{d}{dt} \ln(I-tN) = - (I-tN)^{-1} N g(t)$$ The unique solution of this initial value problem is $$g(t) = I-tN$$ In particular, for $t=1$ we have $\exp(\ln(A)) = A$.
The other direction $\ln(\exp(A))$ doesn't quite make sense yet: you've defined $\ln$ for a matrix whose only eigenvalue is $1$; it would work if all eigenvalues are in the open disk $\{z \in \mathbb C: |z-1|<1\}$, but the eigenvalue of $\exp(A)$ is $e$, and the series $$ \sum_{k=1}^\infty (-1)^k (\exp(A)-I)^k/k $$ will diverge.
It is possible to define a branch of logarithm so that $\ln(\exp(A)) = A$. However, in general one has to be careful, because the logarithm is multivalued. Even for scalars, it's impossible to define a single logarithm function so that $\log(\exp(z)) = z$ for all $z$, because $\exp(2\pi i) = \exp(0) = 1$.