A ring $R$ has no nonzero nilpotent elements Prove that a ring $R$ has no nonzero nilpotent elements if and only if for any $a\in R$, $a^2=0 \implies a=0$.
$"\Rightarrow"$ Assume that $R$ is ring has no nonzero nilpotent elements. Suppose $a^2=0$. If $a\neq 0$, then $a$ is a nilpotent elements of $R$, so $R$ has a nonzero nilpotent, which contradict the assumption. Hence, $a$ must be zero.
$"\Leftarrow"$ Assume that for any $a \in R$, if $a^2=0$ then $a=0.$ Let $a \in R$. Let us assume for the contrary that there is $0 \neq a \in R$ such that $a^n=0$. We will distinguish two cases:
(i) $n=2k+1$, so $a^n=(a^k)^2(a)=0$ which implies that $(a^k)^2=0$ so by assumption $a^k=0$.
(ii) $n=2k$, so we must have $a^k=0$. 
I can not see why $a$ must be zero here. 
I would appreciate any thoughts or other suggestion ways to prove that.
Thank you.
 A: In the proof it has been shown that if $a^{2k}=0$ or $a^{2k+1}=0$ then $a^{k}=0$.
Therefore whatever power of $a$ is zero you can find a smaller power which is zero until you get $a=0$.
A: Here is a quick and simple way to prove this, which does not depend on even/odd cases etc., or any of the (iteratively) linked citings, but is indeed quite concise and self contained:
As pointed out by our OP Ahmed in the text of the question itself, the "$\Rightarrow$" direction is virtually self-evident, that is, if $R$ has no non-zero nilpotents and $a^2 = 0$, then clearly $a = 0$ as well.
As for the "$\Leftarrow$" direction, we observe that 
$a^2 = 0 \Longrightarrow a = 0 \tag 1$
implies, for any $x \in R$ and $m \ge 1$,
$x^{2^m} = 0 \Longrightarrow x^{2^{m - 1}} = 0, \tag 2$
since
$(x^{2^{m - 1}})^2 = x^{2^{m - 1}}x^{2^{m - 1}} = x^{2^{m - 1} + x^{2^m - 1}} = x^{2^1 2^{m - 1}} = x^{2^m} = 0. \tag 3$ 
In light of (2), we may evidently proceed downward, one decrement of $m$ at a time, until we eventually reach $m = 1$, viz:
$x^{2^m} = 0 \Longrightarrow x^{2^{m - 1}} = 0 \Longrightarrow x^{2^{m - 2}} = 0$
$\Longrightarrow \ldots \Longrightarrow x^2 = 0 \Longrightarrow x = 0. \tag 4$
Now if $b \in R$ is nilpotent,
$b^n = 0, \tag 5$
we simply choose $m$ sufficiently large that
$2^m \ge n; \tag 6$
then
$b^{2^m} = b^{2^m - n}b^n = 0; \tag 7$
at this point (2)-(4) take over and we conclude that
$b = 0; \tag 8$
$R$ has no non-zere nilpotent elements.  $OE\Delta$.
