Assume $f(x)=(1+x)^n\in \mathbb F_p[x]$, can we find a general formula for $f(x)$?

Question

We know that $(1+x)^n=\sum_{i=0}^{n}\frac{n!}{i!(n-i)!}x^n \in \mathbb Q[x]$. Assume $f(x)=(1+x)^n\in \mathbb F_p[x]$, when $n>p,$ can we find a general formula for $f(x)$?

Answer 1

$(1+x)^{p^\ell}=\sum_{k=0}^{p^\ell}\binom{p^\ell}{k}x^k$, but for all $1\le k\le p^\ell-1$ we have $\binom{p^\ell}{k}\equiv 0\pmod{p}$ so $$ (1+x)^{p^\ell}=\sum_{k=0}^{p^\ell}\binom{p^\ell}{k}x^k=1+x^{p^\ell} $$ in $\mathbb{F}_p[x]$.

Then for any $n=p^\ell\cdot m$ where $m<p$ we have $$ (1+x)^n=(1+x)^{p^\ell\cdot m}=\left((1+x)^{p^\ell}\right)^m=\left(1+x^{p^\ell}\right)^m=\sum_{k=0}^m\binom{m}{k}x^{kp^\ell}. $$