The group $(1+p\mathbb Z_p)/(1+p^{n}\mathbb Z_p)$ I want know some information about the group 
\begin{equation*}
\frac{(1+p\mathbb Z_p)}{(1+p^{n}\mathbb Z_p)}
\end{equation*}
(the Quotient group).
What is the order of this group? I guess $p^{n-1}$ But how can I prove this?
How does an element of this group look like?
$\mathbb Z_p$ is the ring of the p-adic integers.
 A: The map $1+p^i\mathbb Z_p \to \mathbb Z_p/p\mathbb Z_p,\ 1 + p^ia \mapsto a \bmod p$, has kernel precisely $1+p^{i+1}\mathbb Z_p$, so  $$(1+p^i\mathbb Z_p)/(1+p^{i+1}\mathbb Z_p) \cong \mathbb Z/p\mathbb Z.$$ Hence $(1+p\mathbb Z_p)/(1+p^n\mathbb Z_p)$ is a successive extension of $\mathbb Z/p\mathbb Z$, and indeed has order $p^{n-1}$, as you guessed. This doesn't provide a complete answer though, sorry.
A: The group $U_n:=1+p^n\mathbb{Z}_p$ is the kernel of $U \to (\mathbb Z/p^n \mathbb Z)^*$, the map being the (sujective) projection on the "n-th coordinate". Hence you have the exact sequences :
$$1\to U_1 \to U\to (\mathbb{Z}/ p \mathbb{Z})^*\to 1, \quad (1) $$
$$ 1\to U_n \to U\to (\mathbb{Z}/ p^n \mathbb{Z})^* \to 1, \quad (2)  $$ 
The first sequence gives : 
$$ 1 \to U_1/U_n \to U/U_n \to (\mathbb{Z}/ p \mathbb{Z})^*\to 1,\quad (3)$$
since $U_n\subset U_1$ (isomorphism theorem). 
On the other hand, we know that (using the sequence (2)) that $$\vert U/U_n \vert = \vert (\mathbb{Z}/ p^n \mathbb{Z})^*\vert = \varphi(p^n)=(p-1)p^{n-1}$$ ( $\varphi$ is the Euler ..).
Finaly,the last sequence (3) gives $\vert U/U_n \vert =\vert U_1/U_n \vert\times \vert(\mathbb{Z}/ p \mathbb{Z})^* \vert$. Hence $\vert U_1/U_n \vert=p^{n-1}$. 
PS: Serre's book "A course in Arithmetic" is a very good reference for p-adic numbers.
A: To describe elements of $(1+p\mathbf Z_p)/(1+p^n\mathbf Z_p)$ concretely, observe for $p$-adic units $u$ and $v$ that $|u/v -1|_p = |u-v|_p$, so the multiplicative congruence $u \equiv v \bmod (1+p^n\mathbf Z_p)$ is the same as the additive congruence $u \equiv v \bmod p^n\mathbf Z_p$. Therefore you can think of elements of the quotient group as ordinary congruence classes $1+ a_1p+ a_2p^2 + \cdots + a_{n-1}p^{n-1} \bmod  p^n\mathbf Z_p$ under multiplication, where the $a_i$ are $p$-adic digits.
This lets us easily count the size of the quotient group: each $a_i$ has $p$ values so the size of the group is $p^{n-1}$. (The same reasoning shows $(1+ p^m\mathbf Z_p)/(1+p^n\mathbf Z_p)$ for $1 \leq m \leq n$ has order $p^{n-m}$: coset representatives are $1+ a_mp^m + \cdots + a_{n-1}p^{n-1} \bmod p^{n}\mathbf Z_p$.)   
This does not explain what the group structure is. Is it cyclic? The answer turns out to be yes except when $p = 2$ and $n \geq 3$. The best explanation uses the $p$-adic exponential function, which converges on $p\mathbf Z_p$ for $p \not= 2$ and on $4\mathbf Z_2$ (not $2\mathbf Z_2$!) when $p= 2$.
First assume $p \not=2$. Then the $p$-adic exponential function is a group isomorphism $p\mathbf Z_p \rightarrow 1+p\mathbf Z_p$ that is also an isometry: $|e^x - e^y|_p = |x-y|_p$ for all $x$ and $y$ in $p\mathbf Z_p$. In particular, for any positive integer $n$ we have $|e^x - 1|_p \leq 1/p^n$ if and only if $|x|_p \leq 1/p^n$. Therefore the $p$-adic exponential function maps $p^n\mathbf Z_p$ onto $1+p^n\mathbf Z_p$, so $$p\mathbf Z_p/p^n\mathbf Z_p \cong (1+p\mathbf Z_p)/(1+p^{n}\mathbf Z_p).$$
The group on the left side is isomorphic to $\mathbf Z_p/p^{n-1}\mathbf Z_p$ by division by $p$, and that is cyclic of order $p^{n-1}$. 
What if $p = 2$? In this case the group $(1+2\mathbf Z_2)/(1+2^n\mathbf Z_2)$ is not cyclic for $n\geq 3$ since there is more than one element of order $2$: look at $-1 \bmod 2^n\mathbf Z_2$ and $1+2^{n-1} \bmod 2^n\mathbf Z_2$. (These are different for $n\geq 3$ but they are the same for $n= 1$ or $2$.) Our previous work can be adapted to this case, but we have to move deeper into the group $1+2\mathbf Z_2$. The $2$-adic exponential function does not converge on $2\mathbf Z_2$ but it does converge on $4\mathbf Z_2$ and gives an isomorphism $4\mathbf Z_2 \rightarrow 1+4\mathbf Z_2$ that is a $2$-adic isometry. Since $1+ 2\mathbf Z_2 \cong \{\pm 1\} \times (1+4\mathbf Z_2)$ as (topological) groups, you can develop a result for $p=2$ that is analogous to the case of odd $p$: for $n\geq 2$, which makes $1+2^n\mathbf Z_2 \subset 1+4\mathbf Z_2$, we have $$(1+2\mathbf Z_2)/(1+2^n\mathbf Z_2) \cong \{\pm 1\} \times ((1+ 4\mathbf Z_2)/(1+ 2^n\mathbf Z_2)) \cong  \{\pm 1\} \times (4\mathbf Z_2/2^n\mathbf Z_2),$$
where the second isomorphism comes from the $2$-adic exponential (right to left) or $2$-adic logarithm (left to right). That last group is isomorphic to $\mathbf Z/2\mathbf Z \times \mathbf Z/2^{n-2}\mathbf Z$, which is not cyclic when $n\geq 3$. 
The structure of $(1+2\mathbf Z_2)/(1+2^n\mathbf Z_2)$ when $n = 1$ is trivial.
