Let $p$ be an odd prime number and $n$ be a positive integer. Use the binomial theorem to show that $(1+p)^{p^{n-1}} \equiv 1 \mod p^n$ but $(1+p)^{p^{n-2}} \ne 1 \mod p^n$ Deduce that $(1+p)$ is an element with order $p^{n-1}$ of the multiplicative group $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$
I started with $\sum_{k=0}^{p^{n-1}} {p^{n-1} \choose k} 1^k p^{p^{n-1}-k}$
Writing out terms looks like $(1)p^{p^{n-1}-0} + (1)p^{p^{n-1}-1} + (1)p^{p^{n-1}-2} + ... + (1)p^{p^{n-1} - p^{n-1}}$ is this the correct way to write this sum?
$p^{p^{n-1}} + p^{p^{n-1}}p^{-1} + p^{p^{n-1}}p^{-2} + ... + p^{p^{n-1}}p^{-p^{n-1}}$ reduces to $p^{p^{n-1}}(p^{-1} + p^{-2} + ... + p^{-p^{n-1}})$ somehow evaluates to $1 \mod p^n$
Any help appreciated, thanks!
We have
$$(a+b)^{p^{n-1}}=\sum_{k=0}^{p^{n-1}}\binom{p^{n-1}}ka^{p^{n-1}-k}b^k\implies(1+p)^{p^{n-1}}=\sum_{k=0}^{p^{n-1}}\binom{p^{n-1}}kp^k$$
Now prove (something very similar to this appears in a standard proof of Sylow's first theorem)
Lemma: For $\;m\in\Bbb N\;,\;\;0\le m\le n-1\;$ and $\;0< k< p^{n-1}$ , we have that
$$\;p^m\,\mid\,\left(p^{n-1}-k\right)\;\iff p^m\,\mid\,k$$
Assuming the above, we have that
$$\binom{p^{n-1}}k=\frac{\left(p^{n-1}-(k-1)\right)\left(p^{n-1}-(k-2)\right)\cdot\ldots\cdot p^{n-1}}{k(k-1)(k-2)\cdot\ldots\cdot1}\implies p^{n-1}\,\mid\,\binom{p^{n-1}}k$$
and it follows that $\;\binom{p^{n-1}}k p^k=0\pmod{p^n}\;$ , so we get
$$(1+p)^{p^{n-1}}=\sum_{k=0}^{p^{n-1}}\binom{p^{n-1}}kp^k=\left(1+p^{p^{n-1}}\right)\pmod{p^n }=1\pmod{p^n}$$