Usually, it is done like this: If $[t]$ denotes the greatest integer less than or equal to $t$, then
$$ \text{the exponent of $p$ in $n!$} = \bigg[\frac{n}{p}\bigg] + \bigg[\frac{n}{p^2}\bigg] + \bigg[\frac{n}{p^3}\bigg] + \cdots $$
and it does give the answer, but, it is totally non-intuitive and i can't use it comfortably enough!
On
It is easily understood in terms of double counting: if $\operatorname{ord}_p m$ denotes the exponent of the factor $p$ in $m$ and
$$ \mathbf{1}_{\{\cdots\}} = \begin{cases} 1, & \text{if $\cdots$ is true} \\ 0, & \text{if $\cdots$ is false} \end{cases} $$
denotes the indicator function for the statement $\cdots$, then
$$ \operatorname{ord}_p (n!) = \sum_{k=1}^{n} \operatorname{ord}_p k = \sum_{k=1}^{n} \sum_{l=1}^{\infty} \mathbf{1}_{\{p^l \mid k\}} = \sum_{l=1}^{\infty} \sum_{k=1}^{n} \mathbf{1}_{\{p^l \mid k\}} = \sum_{l=1}^{\infty} \bigg\lfloor \frac{n}{p^l} \bigg\rfloor. \tag{1}$$
The following figure demonstrates the case $n = 65$ with $p=2$.
$\hspace{4em}$
Here, the presence of a black bead at $(k, l)$ means that $p^l$ divides $k$. So
The number of beads at $k$ represents the highest power of $p$ that divides $k$, hence is $\operatorname{ord}_p k$.
The number of beads at height $l$ represents the total number of integers $\leq n$ which are divisible by $p^l$, i.e., the number of multiples of $p^l$ in $\{1, \cdots, n\}$. This is exactly $\lfloor n/p^l \rfloor$.
Then the LHS of $\text{(1)}$ counts the total number of beans from left to right, while the RHS of $\text{(1)}$ counts the total number of beans from bottom to top.
The $\left\lfloor \frac np \right\rfloor$ counts one factor of $p$ from every multiple of $p$. The $\left\lfloor \frac n{p^2} \right\rfloor$ counts one additional factor of $p$ from every multiple of $p^2$ and so on. As an example, take $p=5, n=679$. There are $\left\lfloor \frac {679}5 \right\rfloor=135$ multiples of $5$, starting with $5$ and ending with $675=5\cdot 135$. The multiples of $25$ have another factor of $5$ and there are $\left\lfloor \frac {679}{25} \right\rfloor=27$ of them. The multiples of $5^3=125$ have another factor of $5$ and the one multiple of $5^4=625$ has yet one more.