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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 $f(x)$?

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    Presumably you are looking for [Lucas's theorem](https://en.wikipedia.org/wiki/Lucas's_theorem). That's what you get when you combine the binomial formula with freshman's dream: $(1+x)^{p^\ell}=1+x^{p^\ell}$ in the ring $\Bbb{F}_p[x]$. – 2017-01-24
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    @Jyrki Lahtonen: I didn't know this identity had a name. Thanks for sharing! – 2017-01-24
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    $(1+x)^n=\sum_{i=0}^{n}\frac{n!}{i!(n-i)!}x^n$ stays true in $\mathbb{F}_p[X]$. All you have to do is using that $p = 0$ in $\mathbb{F}_p$ and simplify the *integer* $\frac{n!}{i!(n-i)!}$ accordingly – 2017-01-24
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    @Jyrki Lahtonen, Thanks for response – 2017-01-24

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$(1+x)^{p^\ell}=\sum_{k=0}^{p^\ell}\binom{p^\ell}{k}x^k$, 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