@DeadMG: Suppose
$$
n=\sum_{k\ge0}d_kp^k
$$
Then
$$\begin{align}
\left\lfloor\frac{n}{p}\right\rfloor&=\sum_{k\ge1}d_kp^{k-1}\\
\left\lfloor\frac{n}{p^2}\right\rfloor&=\sum_{k\ge2}d_kp^{k-2}\\
\left\lfloor\frac{n}{p^3}\right\rfloor&=\sum_{k\ge3}d_kp^{k-3}\\
&\vdots\\
\left\lfloor\frac{n}{p^j}\right\rfloor&=\sum_{k\ge j}d_kp^{k-j}\\
&\vdots
\end{align}
$$
So the number of factors of $p$ in $n!$ is
$$
\begin{align}
\sum_{j\ge1}\left\lfloor\frac{n}{p^j}\right\rfloor
&=\sum_{j\ge1}\sum_{k\ge j}d_kp^{k-j}\\