Show that any invertible matrix has a logarithm.

It suffices to show that each $\mathrm{\lambda <!--\; \lambda \; -->}$-Jordan block has a logarithm if $\mathrm{\lambda <!--\; \lambda \; -->}\ne <!--\; \ne \; -->0$
First note that the exponential of a Jordan block is
$\left[\begin{array}{ccccc}\mathrm{\lambda <!--\; \lambda \; -->}& 1& 0& \cdots <!--\; \cdots \; -->& 0\\ & \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& 0\\ & & & \ddots <!--\; \ddots \; -->& 1\\ & & & & \mathrm{\lambda <!--\; \lambda \; -->}\end{array}\right]\stackrel{\exp }{\mapsto <!--\; \mapsto \; -->}\left[\begin{array}{ccccc}{e}^{\mathrm{\lambda <!--\; \lambda \; -->}}& e^{\mathrm{\lambda <!--\; \lambda \; -->}}& \frac{e^{\mathrm{\lambda <!--\; \lambda \; -->}}}{2!}& \cdots <!--\; \cdots \; -->& \frac{e^{\mathrm{\lambda <!--\; \lambda \; -->}}}{(k-<!--\; -\; -->1)!}\\ & e^{\mathrm{\lambda <!--\; \lambda \; -->}}& \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & \ddots <!--\; \ddots \; -->& e^{\mathrm{\lambda <!--\; \lambda \; -->}}& \frac{e^{\mathrm{\lambda <!--\; \lambda \; -->}}}{2!}\\ & & & e^{\mathrm{\lambda <!--\; \lambda \; -->}}& e^{\mathrm{\lambda <!--\; \lambda \; -->}}\\ & & & & e^{\mathrm{\lambda <!--\; \lambda \; -->}}\end{array}\right].$
So, reversing the process, given a Jordan block
$J=\left[\begin{array}{cccc}a& 1& \cdots <!--\; \cdots \; -->& 0\\ & a& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & \ddots <!--\; \ddots \; -->& 1\\ & & & a\end{array}\right]$
with $a\ne <!--\; \ne \; -->0$, we want to show that $J$ is similar to
$\left[\begin{array}{ccccc}a& a& \frac{a}{2!}& \cdots <!--\; \cdots \; -->& \frac{a}{(k-<!--\; -\; -->1)!}\\ & a& \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & \ddots <!--\; \ddots \; -->& a& \frac{a}{2!}\\ & & & a& a\\ & & & & a\end{array}\right],$
since we know how to find the logarithm of the above matrix, and since $\log <!--\; \; -->(UM{U}^{-<!--\; -\; -->1})=U\log <!--\; \; -->(M)U^{-<!--\; -\; -->1}$. Now, since the scalar matrix $aI$ has the same form with respect to any basis, we can neglect this part, and just ask if we can conjugate
$\begin{array}{}\text{(}{M}_1\text{)}& \left[\begin{array}{cccc}0& 1& \cdots <!--\; \cdots \; -->& 0\\ & 0& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & \ddots <!--\; \ddots \; -->& 1\\ & & & 0\end{array}\right]\end{array}$
into
$\begin{array}{}\text{(}{M}_2\text{)}& \left[\begin{array}{ccccc}0& a& \frac{a}{2!}& \cdots <!--\; \cdots \; -->& \frac{a}{(k-<!--\; -\; -->1)!}\\ & 0& \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & \ddots <!--\; \ddots \; -->& a& \frac{a}{2!}\\ & & & 0& a\\ & & & & 0\end{array}\right].\end{array}$
First of all, we can see algebraically that these two matrices should be similar, since $M_2=aN+a{N}^2+\cdots <!--\; \cdots \; -->+{\displaystyle \frac{a}{(k-<!--\; -\; -->1)!}}{N}^{k-<!--\; -\; -->1}$, where $N$ is the elementary nilpotent matrix of size $k$ (here $N=M_1$, incidentally). Taking powers of $M_2$ shows that $M_2^{k-<!--\; -\; -->1}\ne <!--\; \ne \; -->0M_2^k=0$. Therefore $M_1$ and $M_2$ have the same minimal polynomial $x^k$. I claim they also have the same characteristic polynomial, also $x^k$. The only possibility for the invariant factors of $M_1$ and $M_2$ is $1,1,\dots <!--\; \dots \; -->,1,x^k$. Therefore they are similar.
That said, I thought I would tackle the more general problem: Suppose we want to conjugate
$\begin{array}{}\text{(}{M}_3\text{)}& \left[\begin{array}{cccc}0& 1& \cdots <!--\; \cdots \; -->& 0\\ & 0& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & \ddots <!--\; \ddots \; -->& 1\\ & & & 0\end{array}\right]\end{array}$
into
$\begin{array}{}\text{(}{M}_4\text{)}& \left[\begin{array}{ccccc}0& a_{12}& a_{13}& \cdots <!--\; \cdots \; -->& a_{1k}\\ & 0& a_{23}& \cdots <!--\; \cdots \; -->& a_{2k}\\ & & \ddots <!--\; \ddots \; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ & & & \ddots <!--\; \ddots \; -->& a_{k-<!--\; -\; -->1,k}\\ & & & & 0\end{array}\right]\end{array}$
Given that $\mathcal{B}=\{v_1,\dots <!--\; \dots \; -->,v_k\}$ is an ordered basis such that $M_3=[T{]}_{\mathcal{B}\mathcal{B}}$, conjugating $M_3$ into $M_4$ is equivalent to finding an ordered basis $\mathcal{C}=\{w_1,\dots <!--\; \dots \; -->,w_k\}$ such that $[T{]}_{\mathcal{C}\mathcal{C}}=M_4$
Let $v_1,\dots <!--\; \dots \; -->,v_k$ be the basis in $M_1$. Define $w_1,\dots <!--\; \dots \; -->,w_k$ as follows: let $w_1=v_1$. Now suppose
$w_{i-<!--\; -\; -->1}=\underset{1\le <!--\; \le \; -->j\le <!--\; \le \; -->i-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}{c}_j{v}_j$
Then let
$w_i=\underset{1\le <!--\; \le \; -->j\le <!--\; \le \; -->i-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}{a}_{n-<!--\; -\; -->j}{c}_j{v}_{j+1}.$
We need to make sure $w_i$ are linearly independent. I claim they will be iff $M_3$ is similar to $M_4$ iff $M_4^{k-<!--\; -\; -->1}\ne <!--\; \ne \; -->0$. One condition which will guarantee this is:
$\text{All elements on the superdiagonal of }{M}_4\text{ are nonzero.}$
Perhaps someone can come up with some better conditions for it.