Showing no non-zero element is nilpotent in a ring. Suppose that R is a ring in which $a^{2}=0$ implies that a=0
Show that R has no-non-zero nilpotent element
Attempt:
Recall that an element x of a ring R is called nilpotent IF there exists some positive integer n such that $x^{n}=0$

Attempt:
Suppose that R has a non-zero nilpotent element.
Then, $a^{n}=0$ for some $a\neq 0 \in R$ for some positive n.

Here's where I am stuck. If I can show $n=2$, I can arrive at a contradiction. But how do I do so?
 A: Sketch: It is obvious that the smallest possible $n$ must be odd in your set-up because if $n$ were even, then $n=2k$ and $(a^k)^2=0$, so $a^k=0$.
Assume that $n\geq 3$ and consider $a^{n+1}$.  Let $n=2k+1$ for some integer $k$.  This is zero because $a^{n+1}=a^n\cdot a^1=0\cdot a^1=0$.  But then, $(a^{k+1})^2=0$, so, by assumption $a^{k+1}=0$.  This is a contradiction because $k+1<n$.
A specific example: Suppose that $a^3=0$ (and $3$ is the smallest power of $a$ for which $a^n=0$).  Then $a^4=a^3\cdot a^1=0\cdot a^1=0$.  Moreover, $a^4=(a^2)^2=0$, so by assumption, $a^2=0$, a contradiction because this is a smaller power of $a$ which is equal to $0$.
A: It seems to me, it's the powers of 2 for the exponent, i.e. $a^{2^n} = 0$, which give the implication of $a = 0$ only and not just the even numbered exponents. But here's my proof:
By the contrapositive, $a \neq 0 \Longrightarrow a^2 \neq 0 \Longrightarrow a^4 \neq 0 \Longrightarrow a^{2^n} \neq 0$ for all positive integers $n > 1$. If an element $a$ were nilpotent, then $a^m = 0$ for some fixed $m$. Since we can always find an $n$ such that $2^n > m$, choose such an $n$ and we have $0 = a^m \cdot a^{2^n - m} = a^{2^n}$, which is a contradiction. Thus, no nonzero element of $R$ is nilpotent.
