For any prime $p$ and integer $n$, let
$$
n=d_0+d_1p+d_2p^2+d_3p^3+\dots+d_kp^k\tag{1}
$$
where $0\le d_j\lt p$. $(1)$ is the base-$p$ representation of $n$.
The number of multiples of $p$ not greater than $n$ would be
$$
\left\lfloor\frac np\right\rfloor=d_1p^0+d_2p^1+d_3p^2+d_4p^3+\dots+d_kp^{k-1}\tag{2}
$$
The number of multiples of $p^2$ not greater than $n$ would be
$$
\left\lfloor\frac n{p^2}\right\rfloor=\hphantom{d_1p^0+}d_2p^0+d_3p^1+d_4p^2+\dots+d_kp^{k-2}\tag{3}
$$
The number of multiples of $p^3$ not greater than $n$ would be
$$
\left\lfloor\frac n{p^3}\right\rfloor=\hphantom{d_1p^0+d_2p^1+}d_3p^0+d_4p^1+\dots+d_kp^{k-3}\tag{4}
$$
and so forth.
$(2)$ only counts each multiple of $p^2$ once. To count each multiple of $p^2$ twice, we need to add $(3)$. This only counts each multiple of $p^3$ twice. To count each multiple of $p^3$ three times, we need to add $(4)$, and so on.
After adding up $(2)$, $(3)$, $(4)$, and so on, the coefficient of $d_j$ is
$$
p^{j-1}+p^{j-2}+p^{j-2}+\dots+p^0=\frac{p^j-1}{p-1}\tag{5}
$$
Thus, the sum of $(2)$, $(3)$, $(4)$, and so on is
$$
d_0\frac{p^0-1}{p-1}+d_1\frac{p^1-1}{p-1}+d_2\frac{p^2-1}{p-1}+\dots+d_k\frac{p^k-1}{p-1}=\frac{n-\sum d_j}{p-1}\tag{6}
$$
Therefore, the number of factors of $p$ in $n!$ is
$$
\frac{n-\sum d_j}{p-1}\tag{7}
$$
where $\sum d_j$ is the sum of the base-$p$ digits of $n$.
Examples
For $p=2$, $10=1010_{\text{two}}$, so $\sum d_j=2$. There are $\frac{10-2}{2-1}=8$ factors of $2$ in $10!$
For $p=2$, $20=10100_{\text{two}}$, so $\sum d_j=2$. There are $\frac{20-2}{2-1}=18$ factors of $2$ in $20!$
For $p=2$, $30=11110_{\text{two}}$, so $\sum d_j=4$. There are $\frac{30-4}{2-1}=26$ factors of $2$ in $30!$
For $p=2$, $100=1100100_{\text{two}}$, so $\sum d_j=3$. There are $\frac{100-3}{2-1}=97$ factors of $2$ in $100!$
