This has questions comes to me via a proof to Sylow's First Theorem where $G$ is a finite group of order $p^m \cdot k$, the number $p$ is a prime divisor of $|G|$, and $p^m$ is the highest power of $p$ dividing $|G|$.
Why is it that if prime $p$ divides a term $(p^m \cdot k - l)$ in the numerator of ${p^m \cdot k}\choose {p^m}$ then it divides $p^m - l$ in the denominator the same number of times? This part of the proof's aim is to show that ${p^m \cdot k}\choose {p^m}$ is not divisible by $p$.
I created a specific case where $|G|=3^2 \cdot 2$ so that the number of sets with $3^2$ elements is given by $${{3^2 \cdot 2}\choose {3^2}} = \frac{18 \cdot 17 \cdot 16 \cdots 10}{9 \cdot 8 \cdot 7 \cdots 1}$$ Using $p=3$, I see that 3 divides the term $3^2 \cdot 2 - 6 = 18-6 = 12$ in the numerator 4 times and the term $3^2 - 6 = 9-6 = 3$ in the denominator once. Where am I going wrong?
The highest power of $p$ that divides $p^m\cdot k-l$ is the same as the highest power of $p$ that divides $p^m-l$ because $p^m\cdot k-l\equiv p^m-l\pmod{p^m}$. So they have the same remainder on division by $p^m$, and the highest power of $p$ that divides that remainder is the highest power of $p$ that divides each of them.
Remark: About the "going wrong," $3$ does not divide $12$ $4$ times. Here "divides $4$ times" means that successive division can be done $4$ times. The highest power of $3$ that divides $12$ is $3^1$, so successive zero remainder division can only be done once.