Upper Unitriangular Matrices Let $U$ be the group of the upper unitriangular matrices $n$-$n$ over the field of rationals $\mathbb{Q}$. I know that $U$ is nilpotent and torsion-free. It is also radicable? How it can be proved in an elementary way?
Deifnition
A gorup $G$ is said to be a radicable group iff every element of $G$ has an $n$th root in $G$ for all positive numbers $n$. 
 A: The answer is yes, $U$ is radicable. 
The intuition is that a single matrix acts like a variable that satisfies its minimal polynomial. Any matrix in $U$ has minimal polynomial $(u-1)^N$ for some $N \leq n$. In other words, it behaves much like the $R[x]$-module $R[x]/(x-1)^N$, which are just power series centered at $1$ with at most $N$ terms. Here $R$ is the ring from which we draw the matrix entries, so $R=\mathbb{Q}$, though I'll generally just assume it is commutative, associative, and unital, with a few specific integers being invertible.
Newton's formula
The explicit formula: if $u \in U$ and $m \in \mathbb{Z}$ with $\tfrac 1m \in R$, then set $$v = \sum_{i=0}^N \binom{ 1/m}{i} (u-1)^i$$ where $(u-1)^N=0$. Here we assume the binomial coefficients $\binom{ 1/m}{i}$ make sense in $R$, for instance if $N!$ is invertible in $R$, for instance, if $R=\mathbb{Q}$.
By certain logical constructions in $\mathbb{Q}[x] \leq \mathbb{R}$ we can use calculus II to see that $v^m = u$ up to a multiple of $(u-1)^N$, but a multiple of $0$ is $0$, so $v^m = u$ exactly.
Maybe someone else can explain it better. Another version is $v=\exp(\tfrac 1m \log(1+(u-1)))$, but this gives basically the same issue. I'm more familiar with this version, where you just use properties of exponentials and logarithms in $\mathbb{Q}[\![x]\!]$.
