How can prove this $\binom{n}{p}\equiv \left\lfloor\frac{n}{p}\right\rfloor \pmod {p^2}$

Question

Show that if $n \gt p \gt 0$: $$\binom{n}{p}\equiv \left\lfloor\dfrac{n}{p} \right\rfloor\pmod{ p^2}$$ where $p$ is prime. and $$\binom{n}{p}=\dfrac{n!}{(n-p)!p!}$$ This is theorem? True or false? If wrong, can you find $p$ prime and $n$ such that $$\binom{n}{p}\equiv \ ? \mod {p^2}$$ Thank you everyone

Answer 1

As I said in a comment, a very general answer is provided by this article by Andrew Granville.

In this particular case it appears we have $$ \binom{n}{p} \equiv \binom{n_{0}}{p} + p \binom{n_{1}}{p}\pmod{p^{2}} $$ where $$ n = n_{0} + n_{1} p^{2} + \dots. $$ is the decomposition of $n$ in base $p^{2}$.

This can be seen, as in Lucas' theorem by first analyzing $$(1 + x)^{p^{2}} \pmod {p^{2}},$$ and then $$(1 + x)^{n} = (1+x)^{n_{0}} (1+x)^{p^{2} n_{1}} \cdots \pmod{p^{2}}.$$