The Problem: Let $p$ be an odd prime and let $n$ be a positive integer. Use the Binomial Theorem to show that $(1+p)^{p^{n-2}}\not\equiv1$ (mod $p^n$).
Source: Abstract Algebra, $3^{rd}$ edition, Dummit and Foote.
My Attempt: I have shown that $(1+p)^{p^{n-1}}\equiv1$ (mod $p^n$) for the same $p, n$ by showing that each term in the Binomial Expansion of $(1+p)^{p^{n-1}}$ is divisible by $p^n$ (except, of course, the first term). But the same approach obviously fails for the problem at hand. Instead, I tried to make use of $(1+p)^{p^{n-2}}\equiv1$ (mod $p^{n-1}$), to no avail. Finally, assuming $(1+p)^{p^{n-2}}\equiv1$ mod ($p^n$) did not give rise to a contradiction either. I am officially stuck.
There is a relevant post that involves a lemma using $p$-order $Ord_p(x)$, where $Ord_p(x)$ is the largest $N\in\mathbb{Z}$ such that $p^N|x$; but I did not find the final step in the proof for that lemma convincing enough. Other relevant posts such as this one provide tools that are unnecessarily powerful. Hence this post.
Any help would be greatly appreciated.
