By a left simple ring I mean a ring with no proper, nontrivial left ideal. $R$ be such a ring. Let $u(≠0)\in R$. Then $Ru=R$ (since $1u=u≠0$). Now $Ru=${$ru:r\in R$}. So there is an $r\in R$ such that $ru=1$.
Right simplicity of $R$ implies that there is an $s\in R$ such that $us=1$. $(ru)s=r(us) \implies 1s=r1 \implies s=r=u^{-1}.$
So each nonzero element has a multiplicative inverse. Thus $R$ is a division ring.
Is the claim and/or proof correct?