Prove that $(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n}$ but $(1+p)^{p^{n-2}} \not\equiv 1\pmod{p^n}$, deduce $\text{ord}_{p^n}(p+1)=p^{n-1}$ I need some hints for this problem from Dummit and Foote. (edit: added the full question verbatim for context)
Let $p$ be an odd prime and let $n$ be a positive integer.  Use the binomial theorem to show that $(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n}$ but $(1+p)^{p^{n-2}} \not\equiv 1\pmod{p^n}$. Deduce that $1+p$ is an element of order $p^{n-1}$ in the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$.
For the first part I just need to show that $$p^n \;\left|\; \sum\limits_{k=1}^{p^{n-1}}\binom{p^{n-1}}{k}p^k\right.$$ and it seems obvious enough that it should, since I think I can factor a $p^n$ out of each term, but Im having trouble actually proving this.
For the second part I am unsure what to do.
For reference it is problem 21. on page 60 of the third edition. (section 2.3).
Edit Ok I have managed to show the first part.  From the lemma in the post Bill Dubuque linked to, which I could prove easily, if $z\equiv 1$ mod $p^n$ and $z = p+1$ then $z^p \equiv 1$ mod $p^{n+1}$. Since $z^p-1 = (z-1)(1+z+...+z^{p-1})$ and $(1+z+...+z^{p-1})\equiv 1+1+...+1 = p \equiv 0$ mod $p$.
So it follows by induction that $(p+1)^{p^{n-1}} \equiv 1$ mod $p^n$. 
My problem here is that what Ive proven really only shows that $ord_p(z^p-1) \geq ord_p(z-1)+1$ as far as I can tell, so Im still not sure how I can show that $(1+p)^{p^{n-2}} \not\equiv 1\pmod{p^n}$. I think Im missing something here.
Edit II Ok, I snuck a peek at Pete's notes and I see what I was missing. Really the crucial step was writing $z = 1 + xp$, then I could see how to do it.
 A: Hint: use the binomial formula to establish the following fact.
Let $p$ be an odd prime number and $z$ an integer which is $1$ mod $p$.  Then
$\operatorname{ord}_p(z^p − 1) = \operatorname{ord}_p(z − 1) + 1.$
Here, for a nonzero integer $N$, $\operatorname{ord}_p(N)$ is the largest power of $p$ which divides $N$.
(This is actually taken from a result in some notes of mine from an elementary number theory course, but since the OP asked for a hint only, I wish to obey by not posting the link.)
Note: the hint should serve to solve both parts of the question.
Added: Okay, for more help see Lemma 3 of these notes.
A: For $n=1$, $(1+p)^{p^{n-1}} \equiv 1\pmod {p^n}$ is true. So suppose that $n\geq 2$. Now suppose that our induction hypothesis is: $(1+p)^{p^{n-2}} \equiv 1\pmod {p^{n-1}}$.
It follows that there is an integer $k$ such that
\begin{align*}
(1+p)^{p^{n-2}} & =1+kp^{n-1}\tag 1\\
(1+p)^{p^{n-1}}&=(\color{blue}{(1+p)^{p^{n-2}} })^{p} \tag{$1^0$} \\
&  =(\color{blue}{1+kp^{n-1}})^{p}\tag{2}\\
& =1+\sum _{j=1}^{p}\binom{p}{j} (kp^{n-1} )^{j} \tag{$2^0$}\\
 (1+p)^{p^{n-1}} & =1+p^{n-1}\sum _{j=1}^{p}\binom{p}{j} k^{j}( p^{n-1} )^{j-1}\tag 3\\
 & =1+p^{n-1}\left( kp+p^{n-1}\sum _{j=2}^{p} k^{j}\left( p^{n-1}\right)^{j-2}\right)\tag 4\\
 & ={1+p^{n}\left( k+p^{n-2}\sum _{j=2}^{p} k^{j}\left( p^{n-1}\right)^{j-2}\right)}\tag 5
\end{align*}
Note that:
$1.$ $\quad (2)$ follows by using $(1)$ in RHS of $(1^0)$.
$2.$ $\quad (2^0)$ follows by expanding RHS of $(2)$ using binomial theorem.
$3.$ $\quad (3)$ to $(5)$ is algebraic manipulation to bring $p^n$ outside explicitly.
It follows by induction that $\displaystyle ( 1+p)^{p^{n-1}} \equiv \ 1\ \left(\bmod p^{n}\right)$.
