Since $\binom{n}{m} = 0$ for $n < m$ then hereafter we assume that $n \ge m$. Next we use the well-known and easily proved fact that $(x+1)^p \equiv x^p + 1 \pmod{p}$, meaning that each coefficient of the polynomial $(x+1)^p - (x^p + 1)$ is divisible by $p$. Thus,
$$\begin{align*} (x+1)^n &= (x+1)^{a_0+a_1p+\cdots+a_kp^k} \equiv \prod_{i=0}^k (x^{p^i}+1)^{a_i} \pmod{p} \\ &\equiv \prod_{i=0}^k \left( \sum_{j=0}^{a_i} \binom{a_i}{j} x^{jp^i} \right) \pmod{p} \end{align*}$$
The coefficient of $x^m$ in the LHS is
$$(x+1)^n [x^m] = \binom{n}{m}$$
and the coefficient of $x^m$ in the RHS is the coefficient of the following monomial
$$\binom{a_0}{b_0} x^{b_0} \binom{a_1}{b_1} x^{b_1 p} \cdots \binom{a_k}{b_k} x^{b_k p^k} = x^{b_0+b_1p+\cdots+b_kp^k} \prod_{i=0}^{k} \binom{a_i}{b_i}$$
Equating both coefficients, yields
$$\binom{n}{m} \equiv \prod_{i=0}^{k} \binom{a_i}{b_i} \pmod{p}$$