Suppose $a^4=0$, for some $a \in R[x]/(d)R[x]$, then prove that $1-a$ is invertible.
I was thinking since $a^4 = a \cdot a \cdot a \cdot a=0$, this implies that $a$ has to be zero (?) . Now we have that $1-a=1-0=1$, and $1$ is invertible, since $1 \cdot 1 = 1$. Is it really that simple or am I making a logical error somewhere?
In a ring, $ab=0$ does not imply that one of $a,b$ is equal to $0$. For example, consider $2+2$ in the ring $\mathbb{Z}/4\mathbb{Z}$.
To solve the actual problem, try to find some polynomials $P,Q$ so that
$$(1-a)P(a)=1+a^4Q(a)$$
If a ring has no zero divisors, you have that $ab=0$ implies either $a=0$ or $b=0$. In such a case, $a^4=0$ implies $a=0$. However, rings in general do have zero divisors. For example, $2\not=0$ but $2^2=2\cdot 2=0$ in $\mathbb{Z}_4$.
One way to show an element is invertible is to construct an inverse. For $1-a$ where $a^4=0$, our inverse is $1+a+a^2+a^3$. To see this multiply $(1-a)(1+a+a^2+a^3)$ and $(1+a+a^2+a^3)(1-a)$ out. Both give $1$.
This is an old trick based on the geometric series. Recall that $\dfrac{1}{1-x} = 1+x+x^2+x^3+\cdots$ for any real number $|x|<1$. Formally, $(1-x)(1+x+x^2+\cdots)=1$. It then makes sense that $(1-x)(1+x+x^2+x^3)=1$ if $x^4=x^5=\cdots=0$.
