divisibility - What is the highest power of a prime that divides nPr?

I know that the highest power of a prime which divides $n!$ is given by

$$\left[\frac np\right]+\left[\frac n{p^2}\right]+\left[\frac n{p^3}\right]...$$
Where $[x]$ is the greatest integer function.

For $^{n}P_{r}$, which equals $\frac{n!}{(n-r)!}$, I can find the highest power of $p$ dividing $n!$ and $(n-r)!$, and then subtract the two to get the highest power of $p$ dividing $^{n}P_{r}$.

Now is there a better method to do this, where we don't need to find the highest power of $p$ dividing $(n-r)!$? Can we directly find the power of $p$ dividing the product of $r$ consecutive integers?