The $\exp \circ \log$ function acts as the identity on unipotent matrices. I am working through the exercises in "Lie Groups, Lie Algebras, and Representations" - Hall and can't complete exercise 9 of chapter 2 using the provided hint.  
In chapter 2, Hall defines the matrix exponential as
$$e^X = \sum_{m=0}^\infty \frac{X^m}{m!}$$
and the matrix logarithm as 
$$\log A = \sum_{m=1}^\infty (-1)^{m+1}\frac{(A - I)^m}{m}$$
Exercise 9c asks me to show that $\exp(\log(A)) = A$ when $A$ is unipotent.  The hint provided is: Let $A(t) = I + t(A - I)$.  Show that $\exp(\log(A(t))$ depends polynomially on $t$ and that $\exp(\log(A(t))) = A(t)$ for all sufficiently small $t$.  
I have already have a brute force proof that explicitly composes $e$ with $log$ to demonstrates the claim, and I have verified the hint is true, but I would like to understand how to conclude the statement from the hint.

Verification of the hint for interested future readers:
Theorem 2.8 shows that for $||A - I|| < 1$ (Hilbert-Schmidt Norm) that $e^{\log A} = A$.  The matrix in question $A(t)$ has norm $||A(t) - I|| = ||t(A - I)||$ which can be made arbitrarily small for small $t$.
 A: In the hint you have proved that for each element $ij$ of the matrix, the function $t\mapsto(\exp(\log(A(t))))_{ij}$ is a polynomial in $t$, and that there is an $\epsilon>0$ such that this polynomial agrees with $\delta_{ij}+t(a_{ij}-\delta_{ij})$ everywhere on $[0,\epsilon]$. (Here $\delta_{ij}$ is the Kronecker delta, that is, the elements of the matrix $I$).
Two real polynomials that agree on infinitely many points must be the same, so we can conclude that
$$ (\exp(\log(A(t)))_{ij} = \delta_{ij}+t(a_{ij}-\delta_{ij}) $$
for all real $t$.
Now set $t=1$ to conclude that $(\exp(\log(A)))_{ij}=a_{ij}$.
A: Set $A_t = 1 + t(A - 1)$. Since $A$ is unipotent, $\log A_t = a_n t^n + \cdots + a_0$ is a polynomial in $t$. The coefficients $a_n$ commute, since they're polynomials in $A - 1$. Furthermore, since $A$ is unipotent, it follows that the $a_n$ are nilpotent. Hence
$$\exp(\log A_t) = (\exp t^n a_n)\cdots (\exp t a_1) (\exp a_0)$$
is a polynomial in $t$. The result then follows from the hint.
