Showing that $ab=0$ implies $ba=0$ in a ring Question:
Let $n$ be an integer greater than $1$. In a ring in which $x^{n}=x$ for all $x$, show that $ab=0$ implies $ba=0$.

$(ab)^{n}={ab\cdot\cdot\cdot ab}=a(ba\cdot\cdot\cdot ba)b$
there are $(n-1)$ factor of $b\cdot a$
$(ba)^{n-1}=b^{n-1}a^{n-1}=b^{n}b^{-1}\cdot a^{n}a^{-1}=bb^{-1}aa^{-1}$
I have looked at a number of solutions which begins with

&quot;note that $(ba)^{n}$&quot;

which I dislike.
Is there an alternative solution that follows from my attempt?
Thanks in advance.
$ba=(ba)^n=b(ab)a(ba)^{n-2}=b(0)a(ba)^{n-2}=0$
$0 = ab = (ab)^n = b(ab)^n a = (ba)(ba)^n = (ba)^n+1$ implies $ba = 0$. 
Using the result that in any ring with $x^n = x$ for all $x$ in the ring and if $a^m = 0$ for some $a$ in the ring then $a = 0$.