Let $R$ be a commutative ring with $1$, $S$ a subring also with $1$. Suppose $R\setminus S$ contains a nilpotent element. Prove that $R\setminus S$ also contains an invertible element.
Attempt at solution: Let $x\in\ R\setminus S$ be nilpotent, hence $x^{n} =0$ Since the ring is commutative and contains 1, we can apply the binomial theorem: $x^{n}=0=((x-1)+1)^{n}=(\sum_{k=1}^{n}\dbinom{n}{k}(x-1)^{k})+1$
Moving the term to the left and factoring out $(x-1)$ we obtain $(x-1)(\sum_{k=1}^{n}(-1)\dbinom{n}{k}(x-1)^{k-1})=1$
Hence $x-1$ is invertible. To see that it is not in S suppose it was. Since $1\in S, -1\in S$ hence $(x-1)\in S$ hence $x\in S$, a contradiction.
The $(-1)^{n-k}$ should not be there since it's $((x-1)\color{Red}{+1})^n$ you're expanding.
Once you fix that you should be golden.