Prove that $1+p$ is an element of order $p^{n-1}$ in $(\Bbb Z/p^n\Bbb Z)^\times$ using the binomial theorem

Question

Prove that $1+p$ is an element of order $p^{n-1}$ in $(\Bbb Z/p^n\Bbb Z)^\times$, where $p$ is an odd prime and $n$ a positive integer.

Yeah, I'm aware of this: Showing $1+p$ is an element of order $p^{n-1}$ in $(\mathbb{Z}/p^n\mathbb{Z})^\times$ , but I want to construct a proof using the BINOMIAL THEOREM. Since $x^{p^n}=1$ implies $|x|=p^m$ for some $m\le n$, if we can show that $(1+p)^{p^{n-1}}\equiv1 \space(mod \space p^n)$ and $(1+p)^{p^{n-2}}\not\equiv1 \space(mod \space p^n)$, then the result follows.

Let's take an example: let $p=7$, $n=5$
$(1+7)^{7^{5-1}}=(1+7)^{7^4}=1+C^{7^4}_17+C^{7^4}_27^2+C^{7^4}_37^3+C^{7^4}_47^4+C^{7^4}_57^5+..$
$\equiv1+C^{7^4}_17+C^{7^4}_27^2+C^{7^4}_37^3+C^{7^4}_47^4$
$\equiv1+(7^4i_1)7+(7^3i_2)7^2+(7^2i_3)7^3+(7i_4)7^4$
$\equiv0 \space (mod \space 7^5)$ for some integers $i_1,i_2,i_3,i_4$

so it seems I have to show that $p^{n-k}|C^{p^{n-1}}_k$, but I have no idea how to. Can somebody help? Thanks!

Answer 1

you require to show that $p^n$ divides every term of the expansion after the first.

the coefficient of $p^k$ is: $$ B_k = \binom{p^{n-1}}{k} $$ let the power of $p$ dividing this coefficient be $r_k$, so the power of $p$ dividing the term is $r_k+k$ so we need examine only the cases where $k \lt n$

for any prime $p$ and positive integer $N$ the largest power of $p$ dividing $N!$ is: $$ s(N) = \frac{N-\sigma_p(N)}{p-1} $$ where the function $\sigma_p(N)$ is the sum of digits when $N$ is written in base $p$, so the power of $p$ dividing the binomial coefficient $B_k$ is: $$ s(p^{n-1}) - \bigg(s(k)+s(p^{n-1}-k) \bigg) $$ which is equal to: $$ \frac{(\sigma(k) + \sigma(p^{n-1}-k) - 1}{p-1} $$ and evaluates to the number of carries when performing the addition of $k$ and $p^{n-1}-k$ in base-$p$ notation.

$p^{n-1}$ has $n$ digits, and carries can occur at any of the $n-1$ positions after the leading digit. we have $k \lt p^k$ so $k$ has no more than $k$ p-ary digits. Since the sum is a power of $p$ (with all zero digits after the first),there must be carries at each place from digits $k$ to $n-1$, i.e. at least $n-k$ carries.

Hence the binomial coefficient $B_k$ is divisible by (at least) $p^{n-k}$. Together with the power $p^k$ this mean that the terms is divisible by $p^n$

Answer 2

We have $(1+p)^{p^{n-1}}= \sum_{k=0}^{p^{n-1}}{{p^{n-1}} \choose {k}} p^{k}$.

Now $$v_p({{p^{n-1}} \choose {k}} p^{k}) = v_p ({{p^{n-1}} \choose {k}}) + k \ge (n-1-v_p(k)) + k \ge n, \ \forall \ k\ge 1$$ thus $$v_p (\sum_{k=1}^{p^{n-1}}{{p^{n-1}} \choose {k}} p^{k}) \ge n .$$

Therefore, $(1+p)^{p^{n-1}} \equiv 1 \mod{p^{n}}$.

Next, we show that order is precisely $p^{n-1}$. To see this notice that $(1+p)^{p^{n-2}} \equiv 1 + p^{n-1}\mod{p^{n}}$, if $p$ is an odd prime.

Since for $p$ odd,

$$k - v_p(k) \ge 2, \ \forall \ k \ge 2$$

And for $p=2$ the proposition is false. Let $p =2, n=5 $

$$3^{2^3} = 6561 \equiv 1 \mod{2^5}$$

So, $2^3$ is the order of $3= 1+2$ and not $2^4$ as claimed in the proposition.