Let $p>2$ be an odd number and let $n$ be a positive integer. Prove that $p$ divides $${\sum\limits_{r=1}^{p-1}{r^{p^n}}}$$
My Proof: From multinomial expansion, we know that $${(1 + 2 + 3 + ... + p-1)}^{p^n}= 1^{p^n}+ 2^{p^n}+....+ (p-1)^{p^n} + \sum\limits_{k_1+k_2+...+k_{p-1}={p^n}}{{p^n}\choose{k_1,k_2,..,k_{p-1}}}\prod\limits_{1\leq r\leq p-1}r^{p^{k_{t}}} $$, where $k_1,k_2,..,k_{p-1}\neq{p^n}$. So, we may also write , $${\sum\limits_{r=1}^{p-1}{r^{p^n}}}=\left(\small{\frac{p(p-1)}{2}}\right)^{p^n} - \sum\limits_{k_1+k_2+...+k_{p-1}={p^n}}{{p^n}\choose{k_1,k_2,..,k_{p-1}}}\prod\limits_{1\leq r\leq p-1}r^{p^{k_{t}}} $$ Now, $$\left(\frac{p(p-1)}{2}\right)^{p^n} $$ is divisible by $p$ and $$\sum\limits_{k_1+k_2+...+k_{p-1}={p^n}}{{p^n}\choose{k_1,k_2,..,k_{p-1}}}\prod\limits_{1\leq r\leq p-1}r^{p^{k_{t}}}$$ is also divisible by $p$ since multinomial coefficients consist of $p!$. Therefore, $${\sum\limits_{r=1}^{p-1}{r^{p^n}}}$$ is divisible by $p$.
Please tell me if my proof is correct. If yes, then do I need to provide any more details to my proof? If my proof is incorrect, please help me in proving it,please. Thank you! :)
I find your proof correct, but perhaps you want to know that it can be done in a faster, cleaner way.
For every $1\le r\le p-1$ we have $$r^p\equiv r\pmod p$$ and, in fact, if $k-1$ is a multiple of $p-1$, $$r^k\equiv r\pmod p$$
Since $p^n-1$ is a multiple of $p-1$, $$\sum_{j=1}^{p-1}r^{p^n}\equiv\sum_{j=1}^{p-1}r\equiv 0\pmod p$$