Consider the polynomial ring $A[x]$ and $f(x)=\sum_{i=0}^{n}a_ix^i\in A[x]$, where $A$ is a commutative ring with unit. Show that $f$ is a unit in $A[x]$ if and only if $a_0$ is a unit in $A$ and $a_1,\dots, a_n$ are nilpotent.
Consider the formal power series ring $A[[x]]$ and $f(x)=\sum_{i=0}^{\infty}a_ix^i\in A[[x]]$, where $A$ is a commutative ring with unit. Show that $f$ is a unit in $A[[x]]$ if and only if $a_0$ is a unit in $A$.
How to prove these two questions?
For the second one, the only if is clear from $f(x)g(x)=1$ by comparing coefficients of the terms of degree zero. For the if just do long division for series (remember that it is like long division for polynomials but organizing the terms by increasing degrees).
For the first one, for the if part, consider a large power of $a_0^{-1}f(x)$.
For the only if it is clear $a_0$ is going to be a unit by comparing coefficients in $f(x)g(x)=1$.
For the only if part we can assume $a_0=1$ (by considering $a_0^{-1}f(x)$) to have to write less. We use induction in the number of terms. If $B$ is a nilpotent term and $f(x)$ is a unit then $f(x)-B$ is also a unit.
Begin from polynomials of at most degree one (we can always assume the inverse with the same number of terms by adding if necessary terms with zero coefficients). $f(x)=1+a_1x+a_2x^2+...+a_nx^n$ and $g(x)=1+b_1x+b_2x^2+...+b_nx^n$. From $f(x)g(x)=1$ you can solve, comparing coefficients, for the coefficients of $g$.
For example, $b_1+a_1=0$. Then $b_1=-a_1$. But then $0=b_1a_1+a_2+b_2=-a_1^2$. So, $a_1$ is nilpotent. ...
Hint: If $a$ is nilpotent, and $b$ is a unit, then $ b-a$ is a unit.
Proof: Suppose $a^n = 0$ and $bc = 1$. Then $c(b-a)(1 + (ca) + \ldots + (ca)^{n-1}) = 1- (ca)^n = 1$
