Lie Groups question from Brian Hall's Lie Groups, Lie Algebras and their representations. In page 60 of Hall's textbook, ex. 8 assignment (c), he asks me to prove that if $A$ is a unipotent matrix then $\exp(\log A))=A$.
In the hint he gives to show that for $A(t)=I+t(A-I)$ we get 
$$\exp(\log A(t)) = A(t) , \ t<<1$$
I don't see how I can plug here $t=1$, this equality is true only for $t \rightarrow 0$, right?

If $A$ is unipotent, then $A - I$ is nilpotent, meaning that $(A-I)^n = 0$ for all sufficiently large $n$.  This will turn your power series into a polynomial.
