Lucas’s theorem

Let $m,n\in \mathbb{N}-\{0\}$ be two natural numbers . If $p$ is a prime number and :

$$m=a_k{p}^k+a_{k-1}{p}^{k-1}+\mathrm{\cdots }+a_{1}p+a_0,n=b_k{p}^k+b_{k-1}{p}^{k-1}+\mathrm{\cdots }+b_{1}p+b_0$$

are the base-p expansions of $m$ and $n$ , then the following congruence is true :

$$\left(\genfrac{}{}{0pt}{}{m}{n}\right)\equiv \left(\genfrac{}{}{0pt}{}{a_0}{b_0}\right)\left(\genfrac{}{}{0pt}{}{a_1}{b_1}\right)\mathrm{\cdots }\left(\genfrac{}{}{0pt}{}{a_k}{b_k}\right)(\mathrm{𝚖𝚘𝚍𝚙})$$

Note : the binomial coefficient is defined in the usual way , namely :

$$\left(\genfrac{}{}{0pt}{}{x}{y}\right)=\frac{x!}{y!(x-y)!}$$

if $x\ge y$ and $0$ otherwise (of course , x and y are natural numbers).