Originally Posted by

**TheEmptySet** This problem is from Abstract Algebra 3rd edition Dummit and Foote.

Let p be an odd prime and n a positive integer. Use the Binomial Theorem to show that:

$\displaystyle (1+p)^{p^{n-1}} = 1 (\mbox{mod } p^n)$

but

$\displaystyle (1+p)^{p^{n-2}} \ne 1 (\mbox{mod } p^n)$

Deduce that that $\displaystyle 1+p$ is an element of order $\displaystyle p^{n-1}$ in the multiplicative group $\displaystyle (\mathbb{Z}/p^n\mathbb{Z} \mbox{ }\cdot)$

Thanks

As hint goes by the binomail theorem I get

$\displaystyle (1+p)^{p^{n-1}}=1+\sum_{i=1}^{p^{n-1}}\binom{p^{n-1}}{i}p^i=1+\sum_{i=1}^{p^{n-1}}\left(\frac{p^{n-1}!}{(p^{n-1}-i)!(i)!} \right)p^i$

What I have been trying from here (with no sucess) is to show that $\displaystyle p^{n-1}|\left(\frac{p^{n-1}!}{(p^{n-1}-i)!(i)!} \right)p^i$