How to prove this $p^{j-\left\lfloor\frac{k}{p}\right\rfloor}\mid c_{j}$

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Let $p$ be a prime number and $g\in \mathbb{Z}[x]$.

Let $$\binom{x}{k}=\dfrac{x(x-1)(x-2)\cdots(x-k+1)}{k!} \in \mathbb{Q}[x]$$ for every $k \geq 0$.

Fix an integer $k$. Write the integer-valued polynomial $\binom{g(px)}{k}$ in the form $$\binom{g(px)}{k}=\sum_{i=0}^{\infty}c_{i}\binom{x}{i}$$ for some fixed $c_0, c_1, c_2, \ldots \in \mathbb{Z}$, where all but finitely many $i\geq 0$ satisfy $c_i \neq 0$.

show that $$p^{j-\left\lfloor\frac{k}{p}\right\rfloor}\mid c_{j},\text{where } j=0,1,2,\ldots$$