P. Erdos and J. L. Selfridge proved in the paper THE PRODUCT OF CONSECUTIVE INTEGERS IS NEVER A POWER (click here), that the equation $(n + 1) \cdots(n + k)=x^l \cdots (1)$ has no solution in integers with $k > 2, l > 2, n > 0$. There is a lemma $2$ on page 294, 295 -
LEMMA 2. By deleting a suitably chosen subset of $\pi(k-1)$ of the numbers $a_i (1 \leq i \leq k)$, we have $$a_{i_l}...a_{i_k'},|(k -1)!\cdots (9)$$ where $k’=k -\pi(k-1)$.
For each prime $p < k-1 $ we omit an $a_m$ for which $ n + m$ is divisible by $p$ to the highest power. If $1 \leq i \leq k$ and $i \neq m$, the power of $p$ dividing $n +i$ is the same as the power of $p$ dividing $i-m$. Thus $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)!$.
I understand $p | i-m$ but how does $p^\alpha||a_{i_l}...a_{i_k'}$, implies $p^\alpha|(k-m)!(m-1)! \:\:$ ?
Edit:
Here, $n + i= a_ix^l$, where $a_i$ is $l^{th}$-power free and all its prime factors are less than $k$ (supposing Theorem 2 is false, see page 293 of the paper (click here) for the definition).