Embedding torsion units of an order into torsion units of the reduced order. Let $A$ be an order, i.e. a commutative ring of which the additive group is isomorphic to $\mathbb{Z}^n$ for a certain non-negative integer $n$. Show that there exists an embedding
$$A^{\times}_{\text{tor}}\ \hookrightarrow\ (A_{\text{red}})^{\times}_{\text{tor}},$$
where $A^{\times}_{\text{tor}}$ is the group of torsion units of $A$, and $A_{\text{red}}=A/\sqrt{0_A}$ is the reduced ring of $A$.
Edit: In response to a reply which seems to have been removed; I understand that the quotient map $A\rightarrow A_{\text{red}}$ restricts to a group homomorphism $A^{\times}_{\text{tor}}\rightarrow(A_{\text{red}})^{\times}_{\text{tor}}$, but I am unable to show that this map is injective.
 A: Since this is homework, let me give a hint:
If $a$  lies in the kernel of your map, show that you may write $a = 1 + x$
for some nilpotent $x$.  Now write $a^n = 1$ for some $n$, deduce a corresponding
equation involving $x$, and see where it leads.
Added: Since the OP has now solved the question, let me sketch the answer, based
on the discussion in the comments:
$(1+x)^n = 1$ implies that $nx + $ higher order terms in $x = 0$.  From this it is easy to deduce that $x = 0,$ given that $x$ is nilpotent.  (The OP's gives the following very succinct approach: we may factor out $x$ in the above equation to get $x (n + $ terms involving positive powers of $x) = 0$, and the parenthetical factor is a non-zero divisor in $A$ (since $n$ is not a zero-divisor, and $x$ is nilpotent).)
As an aside, note that
the assumption that $A$ is torsion-free as an abelian group (this is what is really used; of course it follows directly from the assumption that $A$ is an order) is crucial.  There are char. $p$ examples where the given map is not injective.  One of the simplest is obtained by taking $A = \mathbb F_p[x]/(x^p).$ 
