$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

For the rightward implication we can use properties of a ring to see the following $$\begin{array}{cccc}& ra& amp;=rb⟺& \text{<span class="tml-eqn"></span>}\\ & ra-rb& amp;=0⟺& \text{<span class="tml-eqn"></span>}\\ & r(a-b)& amp;=0⟺& \text{<span class="tml-eqn"></span>}\end{array}$$

Therefore we conclude that either $a-b=0$ or $r=0$, the latter of which is impossible through our assumption that $r\ne 0$, therefore $a=b$.

We move on to the leftward implication where our goal is to show that $R$ is a domain, we do so by chasing it's corollary. So let $a,b\in R$ and assume that $ab=0$, note that at this point if $a=0$, our proof is over, therefore we assume that $a\ne 0$, note that we can rewrite $ab=0$ to $ab=a0$, then by our assumption we can conclude that $b=0$, which proves it.

If $x=0_R$ then our proof is done, so instead let's assume that $x\ne 0_R$, we know that it's nilpotent, considering $\{a\in {\mathbb{N}}_1:x^a=0_R\}\subseteq {\mathbb{N}}_0$ it has a least element (as it's non-empty) say $l$ such that $x^l=0_R$, note that $l>1$ because $x^1\ne 0_R$

In that case we can see $x^l=x^{l-1}{x}^1$, since $l$ was the smallest natural number such that $x^l=0_R$ then we must have $x^{l-1}\ne 0_R$ since $l-1\in {\mathbb{N}}_0$. Thus we can conclude that $x{x}^{l-1}=0_R$ where $x,x^{l-1}\ne 0_R$ so that by definition $x$ is a zero divisor.

Since $x$ is nilpotent we have some $n\in {\mathbb{N}}_1$ such that $x^n=0_R$, since $R$ is commutative, then it can be shown by induction that an equation of the form ${\left(ab\right)}^n=a^n{b}^n$, in other words ${\left(rx\right)}^n=r^n{x}^n=r^n{0}_R=0_R$ which proves that $rx$ is nilpotent

If $x^1=0_R$, it clearly holds so lets assume that $x^n=0_R$ for some $n>1$.

We claim that $$v:=1-x+x^2-x^3+...+{(-1)}^{n-1}{x}^n-1=\sum _{i=1}^{n-1}{(-1)}^i{x}^i$$ is $1+x$'s multiplicative inverse.

To see this true, we use distributivity, $$(1+x)\left(\sum _{i=1}^{n-1}{(-1)}^i{x}^i\right)=\sum _{i=1}^{n-1}{(-1)}^i{x}^i+\sum _{i=1}^{n-1}{(-1)}^i{x}^{i+1}$$ Define the following $$S_1\left(k\right):=\sum _{i=1}^{k-1}{(-1)}^i{x}^i=1-x+x^2-x^3+...+{(-1)}^{k-1}{x}^{k-1}$$ and that $$S_2\left(k\right):=\sum _{i=1}^{k-1}{(-1)}^i{x}^{i+1}=x-x^2+x^3-x^4+\dots {(-1)}^{k-2}{x}^{k-1}+{(-1)}^{k-1}{x}^{k-1}$$

Note that $S_2\left(n\right)=S_2(n-1)$ because $S_2\left(n\right)$'s last term is ${(-1)}^{n-1}{x}^n=0_R$ since $x^n=0_R$ .

Through finite induction, we can show that for any $j\in [2\dots n],S_1\left(j\right)+S_2(j-1)=1$ which allows us to conclude that $S_1\left(n\right)+S_2(n-1)=1$ and then $S_1\left(n\right)+S_2\left(n\right)=1$ from the previous paragraph.

Finally this is good because now we continue the chain of equalities we mentioned at distributivity $$\begin{array}{cccc}& (1+x)\left(\sum _{i=1}^{n-1}{(-1)}^i{x}^i\right)& amp;=\sum _{i=1}^{n-1}{(-1)}^i{x}^i+\sum _{i=1}^{n-1}{(-1)}^i{x}^{i+1}& \text{<span class="tml-eqn"></span>}\\ & & amp;=S_1\left(n\right)+S_2\left(n\right)& \text{<span class="tml-eqn"></span>}\\ & & amp;=1& \text{<span class="tml-eqn"></span>}\end{array}$$ Therefore $1+x$ is a unit, as needed.