Login
unity plus nilpotent is unit
Since $x$ is nilpotent, there is a positive integer $n$ such that $x^n=0$ . We multiply $1\!+\!x$ by another ring element: \begin{eqnarray*} (1\!+\!x)\cdot\sum_{j=0}^{n-1}(-1)^jx^j &=& \sum_{j=0}^{n-1}(-1)^jx^j\!+\!\sum_{k=0}^{n-1}(-1)^kx^{k+1}\\ &=& \sum_{j=0}^{n-1}(-1)^jx^j\!-\!\sum_{k=1}^n(-1)^kx^k\\ &=& 1\!+\!\sum_{j=1}^{n-1}(-1)^jx^j\!-\!\sum_{k=1}^{n-1}(-1)^kx^k\!-\!(-1)^nx^n\\ &=& 1\!+\!0\!+\!0\\ &=& 1 \end{eqnarray*} (Note that the summations include the term $(-1)^0x^0$ , which is why $x=0$ is excluded from this case.)
The reversed multiplication gives the same result. Therefore, $1\!+\!x$ has a multiplicative inverse and thus is a unit. 
Note that there is a similarity between this proof and geometric series: The goal was to produce a multiplicative inverse of $1\!+\!x$ , and geometric series yields that
$$\displaystyle \frac{1}{1\!+\!x}=\sum_{n=0}^{\infty} (-1)^nx^n,$$
provided that the summation converges. Since $x$ is nilpotent, the summation has a finite number of nonzero terms and thus converges.