Strange algebra question - with permutations. 
Let $R$ be a reduced ring (i.e. a ring having no nonzero nilpotent elements). Prove that if $a_1, ..., a_n ∈ R$ are such that $a_1a_2 \cdots a_n = 0$, then $a_{\tau(1)}a_{\tau(2)} \cdots a_{\tau(n)} = 0$, for any permutation $\tau$ in the symmetric group $S_n$.

That exercise seems very easy but surprisingly it is not for me. 
What I got so far:
– it is enough to prove that for any transposition $(ij)$ because $S_n$ is generated by all transpositions.
– I've tried to assume contradiction and create a nonzero nilpotent element – no effect so far.
But:
 – I thought also about prove by induction – assuming it works for $n$, I can "catch" any two elements which don't move in my transposition $(ij)$ for example $a_ {n-1}*a_{n}$ and call them "beta" or sth and use induction assumption for $a_1*a_2...*a_{n-2}*beta$ and then rewrite it with disrupting beta? I think it is ok.
Even if I'm right with the induction idea, I would like to see if it is possible to solve it with my first idea – to create a  nonzero nilpotent element.
 A: Result 1. 
If $ab=0$ then $(ba)^2=b(ab)a=0$ and therefore $ba=0$. The theorem is true for $n=2$.
Result 2. 
If $abc=0$ then repeated application of Result 1 gives $bca=cab=0$. Then for any $d$, $dcab=0$ and so $bdca=0$. Then for any $e$, $ebdca=0$ and therefore $aebdc=0$.
Now consider $(acb)^2=acbacb=[a(c)b(a)c]b$. Since $[a(c)b(a)c]=0$ (by $aebdc=0$, applied to $e=c$ and $d=a$) we have $(acb)^2=0$ and therefore $acb=0$. Then $bac=cba=0$ and the theorem is true for $n=3$. 
General proof
It is sufficient to prove the theorem for any transposition. Let $a_1a_2 . . . a_n = 0$ be written as $XaYbZ=0$ where we wish to interchange $a$ and $b$. Applying Results 1 and 2 without further mention we have the following. In each step bracketed expressions are being considered as a single element.
$$XaYbZ=0$$
$$\implies X(aYbZ)=0$$
$$\implies (aYbZ)X=0$$
$$\implies (aY)b(ZX)=0$$
$$\implies b(aY)(ZX)=0$$
$$\implies ba(YZX)=0$$
$$\implies ab(YZX)=0$$
We can now reverse the first 5 steps (with $a$ and $b$ interchanged) to obtain
$$XbYaZ=0$$
and we are finished.
