Show that $f$ is a unit in $R[x] \iff a_0$ is a unit in $R$ and $a_1, \dots, a_n$ are nilpotent. 
Let $R[x]$ be the ring of polynomials in an indeterminate $x$ with coefficients in a ring $R$. Let $f = a_0 + a_1x + · · · + a_nx^n \in R[x]$ where $a_i \in R$. Show that $f$ is a unit in $R[x] \iff a_0$ is a unit in $R$ and $a_1, \dots, a_n$ are nilpotent.

Suppose that $f$ is an unit. Then there exists $g=b_mx^m+\dots +b_0$ such that $fg=1$ that is $$(a_0 + a_1x + · · · + a_nx^n)(b_0+ b_1x+\dots+b_mx^m) = 1$$ since the  rhs doesn't contain any terms with $x$ we must have that $a_0b_0=1$.
How can I proceed to show that the other terms are nilpotents? I was given a hint to compare coefficients, but I don't really understand how to do that in this case.
Multiplying the equations I'll get something of form $$a_nx^nb_mx^m + \dots + a_0b_mx^m+ \dots +a_0b_0 = a_nb_mx^{n+m} + \dots +a_0b_mx^m+ \dots +a_0b_0=1$$ but what can I do with the coefficients?
 A: This property is true when the ring is commutative. I don't think it remains true if you consider a general ring (or, rather, I don't know any proof in the non-commutative case ; if someone know about this case, I'm interested).
First, you can note that : $a_n b_m = 0$ by considering the term of order $n+m$.
Then, you have at the order $n+m-1$ : $a_n b_{m-1} + a_{n-1} b_m = 0$. Multiplying by $a_n$ and using the previous equality : $a_n^2 b_{m-1} = 0$.
At the next step : $a_n b_{m-2} + a_{n-1} b_{m-1} + a_{n-2} b_m = 0$ and by multiplying by $a_n^2$, $a_n^3 b_{m-2} = 0$.
Iteratively, you therefore have : $b_{m+1-k} a_n^k = 0$. (Try to do the induction properly.)
At the final step, $b_0 a_n^{m+1} = 0$. But $a_0 b_0 = 1$ so $b_0$ is invertible, and therfore $a_n^{m+1} = 0$. This proves that $a_n$ is nilpotent.
Now, by noting that the sum of a nilpotent and an invertible remains invertible (here you use again the commutativity), you can procede by induction to prove that each coefficient is nilpotent.
