Torsion-free nilpotent Group I am looking for  a short proof to the following fact: 

In torsion-free nilpotent group we have: an non-trivial element  cannot be conjgate to it inverse.

I know very little about nilpotent torsion-free group,one of their properties is that they have unique roots; i.e. for all $n \ge 1$, $x^n=y^n \Rightarrow x=y$.
Thanks in advance to anyone who is willing to help :-).

Let $G$ be a torsion-free group, and suppose that some non-trivial element $x \in G$ is conjugate to its inverse. So there exists $y \in G$ with $y^{-1}xy=x^{-1}$. 
Let $H = \langle x,y \rangle$ be the subgroup of $G$ generated by $x$ and $y$. Then $y^2$ centralizes both $x$ and $y$, so $y^2 \in Z(H)$. Since $y$ does not centralize $x^k$ for any $k \ne 0$, $x^k$ cannot be a power of $y$, and hence $\langle x \rangle \cap \langle y \rangle = 1$.
Hence $Q = H/\langle y^2 \rangle$  is generated by elements $\bar{x},\bar{y}$ of orders $\infty$ and $2$ with $\bar{y}^{-1} \bar{x} \bar{y} = \bar{x}^{-1}$, so $Q$ is the infinite dihedral group, which is not nilpotent, because it has trivial centre.
So $G$ is not nilpotent.
