Given a matrix $A$ which has the form of a Jordan normal matrix:
$$ A = \left(\begin{array}{cccc} \log(\phi_1) & 1& ...& 0\\ & \log( \phi_2)& & \\ & & \ddots & 1 \\ 0 & & & \log (\phi_n) \\ \end{array}\right)$$
how can we prove that the Jordan normal form of $ \exp(A)= VJV^{-1} $ is $J$:
$$ J = \left(\begin{array}{cccc} \phi_1 & 1& ...& 0\\ & \phi_2& & \\ & & \ddots & 1 \\ 0 & & & \phi_n \\ \end{array}\right)$$
that has the exact same block form as $A$, but only with the exponential of the eigenvalues?
EDIT:
I tried the $2 \times 2$ case to see if it worked and this is what I got:
$$ A = \left(\begin{array}{cc} \log(\phi_1) & 1\\ 0 & \log( \phi_2) \\ \end{array}\right)$$
We can separate this into to $2$ matricies:
$$ A = \left(\begin{array}{cc} \log(\phi_1) & 0\\ 0 & \log( \phi_2) \\ \end{array}\right) + \left(\begin{array}{cc} 0 & 1\\ 0 & 0 \\ \end{array}\right) $$
Exponentiating:
$$ e^A = \left(\begin{array}{cc} \phi_1 & 0\\ 0 & \phi_2 \\ \end{array}\right) \times \left[I + \left(\begin{array}{cc} 0 & 1\\ 0 & 0 \\ \end{array}\right) \right] = \left(\begin{array}{cc} \phi_1 & \phi_1\\ 0 & \phi_2 \\ \end{array}\right)$$
Now, its eigenvalues have all algebraic multiplicity of 1 so the Jordan normal form of $e^A$ is:
$$ J = \left(\begin{array}{cc} \phi_1 & 0\\ 0 & \phi_2 \\ \end{array}\right)$$
which is not quite what we wanted, I dont know what to do.
Hint: If $J$ is the $n\times n$ Jordan block with zero eigenvalue, compute the kernel of $exp(J)$.