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.