I want to solve the following exercise from Dummit & Foote's Abstract Algebra:
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 $\left( \mathbb{Z}/p^n \mathbb{Z} \right)^\times$.
I have managed to show that $$(1+p)^{p^{n-1}} \equiv 1 \pmod{p^n} $$ for all primes $p$ (even $p=2$), and for all $n \in \mathbb{Z}^+$.
However, I got stuck while trying to prove that $$(1+p)^{p^{n-2}} \not \equiv 1 \pmod{p^n} $$ (obviously, this statement makes sense only for $n \geq 2$, and here the requirement that $p$ is odd is essential). I've expanded the LHS using the binomial theorem, and I counted the powers of $p$ in each term. My proof will be complete provided that the following holds for all odd primes $p$, integers $n \geq 3$ and $2 \leq k \leq p^{n-2}$:
$$\sum_{m=1}^\infty \left\lfloor \frac{k}{p^m} \right\rfloor+\left\lfloor \frac{p^{n-2}-k}{p^m} \right\rfloor=\frac{p^{n-2}-1}{p-1}+2-n $$
Can anyone help me prove this? Alternatively, can anyone provide me with another solution of the problem?
Thanks.
Your identity does not always hold. For example, for $p = 5, n = 5$ and $k = 5$, the right hand side is
$$\frac{5^3-1}{4} - 3 = 31-3 = 28,$$
but the left is
$$\left\lfloor \frac{5}{5}\right\rfloor + \left\lfloor \frac{120}{5}\right\rfloor + \left\lfloor \frac{120}{25}\right\rfloor = 1 + 24 + 4 = 29.$$
Suggestion for an alternative proof: For an odd prime $p$ and $k\in\mathbb{N}$ we have
$$(1+p)^{p^k} \equiv 1 + p^{k+1} \pmod{p^{k+2}}.$$
That is easily proved by induction.