TITLE: Show that $(\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}$
QUESTION [7 upvotes]: Show that $(\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}$.

Anyway how and what method is used to prove this.
I still have no idea now.
really thanks for your help

REPLY [2 votes]: Key observation #1:
If $n$ is a nilpotent element of a ring, then the geometric sum formula
$$
\frac1{1-n}=1+n+n^2+n^3+\cdots
$$
shows that $1-n$ is invertible.
Key observation #2:
In your ring the constant term of the product of cosets of $a_0+a_1x+a_2x^2+\cdots$ and $b_0+b_1x+\cdots$ is $a_0b_0$.
[Edit:] The point of these observations was to identify the set of units of
your ring. Compare with Martin Brandenburg's comment! [/Edit]
Key observation task #3:
What's the multiplicative group generated by the cosets of $1+x$, $1+x^2$, $\ldots$ , $1+x^n$? Can you show that this is free abelian? Look at the low degree terms!
[Edit:]
Hint: If $u=1+a_1x+a_2x^2+\cdots\in 1+(x)$, then show that
$$
(1+x)^{b_1}u\in 1+(x^2)
$$
if and only if $b_1=-a_1.$
After that try to show that there is a unique integer $b_2$ with the property that
$$
(1+x^2)^{-b_2}(1+x)^{-a_1}u\in 1+(x^3).
$$
Proceed recursively.
[/Edit]