Prove $(1+p)^{p^{n-2}} = 1+p^{n-1}+up^n$ for some $u$

Question

I am trying to prove exactly what the title says $(1+p)^{p^{n-2}} = 1+p^{n-1}+up^n$ for some $u$, where $p$ is an odd prime and $n$ is a positive integer. The hint is to use the binomial theorem, so $$(1+p)^{p^{n-2}} = \sum_{k =0}^{n-2}{{p^{n-2}}\choose{k}}p^k$$ Which I was playing with and I see whats going on. Basically the first and the last terms get us the first two pieces, and to get the final term of our equation we need to show that $$p^{n-k}\vert{{p^{n-2}}\choose{k}}$$ for $2\leq k< n-2$, and then the $p^k$ gives us the remaining prime powers to show that each of these terms is a multiple of $p^n$. At least I think that is the idea. I'm just not sure how to prove the divisibility portion. I know $p\vert {{p}\choose{k}}$, so possibly induction on this?

I could just use some pointers. Thanks in advance.

Answer 1

$$(1+pk)^p\equiv1+p(pk)+\binom p2(pk)^2\equiv1+p^2k\pmod{p^{1+2}}$$

Let $$(1+pk)^{p^m}=1+kp^{m+1}+cp^{m+2}$$

$$(1+pk)^{p^{m+1}}=\left((1+pk)^{p^m}\right)^p=(1+p^{m+1}(k+cp))^p=1+p^{m+2}(k+cp)+\binom p2(k+cp)^2p^{2m+2}+\cdots\equiv1+kp^{m+2}+c'p^{m+3}$$ if $2m+2+1\ge m+3\iff m\ge0$

Answer 2

There is a useful theorem stating that the valuation of $\binom{p^n}{k}$ at $p$ is $n-v(k)$ for $k\geq1$; see, for instance, this article: https://perso.univ-rennes1.fr/matthieu.romagny/notes/p_adic_formulas.pdf (part 5 of the lemma). This should give you a good start.