$a+1,a-1$ invertible for nilpotent element
You know that if $a$ is nilpotent, then $1-a$ is invertible. Hence, also $-(1-a)=a-1$ is invertible. Hence, since $-a$ is also nilpotent, also $1-(-a)=1+a$ is invertible.
If $a$ is nilpotent, then $-a$ is nilpotent. Note that you start from $a^n=0$, not $a^n=1$.