proof of Lucas’s theorem by binomial expansion

We work with polynomials in $x$ over the integers modulo $p$.
Bythe binomial theorem we have ${(1+x)}^p=1+x^p$. Moregenerally, by induction on $i$ we have ${(1+x)}^{p^i}=1+x^{p^i}$.

Hence the following holds:

$${(1+x)}^n={(1+x)}^{\left[{\sum }_{i=0}^k{a}_i{p}^i\right]}=\prod _{i=0}^k{(1+x^{p^i})}^{a_i}=\prod _{i=0}^k\sum _{b=0}^{a_i}\left(\genfrac{}{}{0pt}{}{a_i}{b}\right)x^{b{p}^i}$$

Then the coefficient on $x^m$ on the left hand side is $\left(\genfrac{}{}{0pt}{}{n}{m}\right)$.

As $m$ is uniquely base $p$, the coefficient on $x^m$on the right hand side is ${\prod }_{i=0}^k\left(\genfrac{}{}{0pt}{}{a_i}{b_i}\right)$.

Equating the coefficients on $x^m$ on either therefore yieldsthe result.