Prove that $\dfrac{b^{n-1}a(a+b)(a+2b)\cdots(a+(n-1)b)}{n!}$ is an integer 
Let $a$ and $b$ be integers and $n$ a positive integer. Prove that $$\dfrac{b^{n-1}a(a+b)(a+2b)\cdots(a+(n-1)b)}{n!}$$ is an integer.

Define $v_p(x)$ such that if $v_p(x) = n$, then $p^n \mid x$ but $p^{n+1} \nmid x$. Then we need to show that $v_p(b^{n-1})+v_p(a)+v_p(a+b)+\cdots+v_p(a+(n-1)b)\geq v_p(n!)$ for all primes $p$. How should we do that?
 A: Fix a prime natural number $p$.  For an integer $m$ and a nonnegative integer $r$, let $p^r\parallel m$ denote the condition that $p^r\mid m$ but $p^{r+1}\nmid m$.  We shall ignore the trivial cases (namely, $n=1$, $a=0$, and $b=0$).  Suppose that $p^k\parallel a$ and $p^l\parallel b$ for some integers $k,l\geq 0$.
First, we assume that $k <  l$ (whence $l\geq 1$).  It follows immediately that $v_p(a+jb)=v_p(a)=k$ for all $j=0,1,2,\ldots,n-1$.  Thus, $$v_p\left(b^{n-1}\,\prod_{j=0}^{n-1}\,(a+jb)\right)=v_p\left(b^{n-1}\right)+\sum_{j=0}^{n-1}\,v_p(a+jb)=(n-1)l+nk\,.$$  Note that $$v_p(n!)=\sum_{j=1}^\infty\,\left\lfloor\frac{n}{p^j}\right\rfloor<\sum_{j=1}^\infty\,\frac{n}{p^j}\leq \sum_{j=1}^\infty\,\frac{n}{2^j}=n\,.$$
Consequently, $\displaystyle v_p(n!)\leq n-1\leq (n-1)l\leq (n-1)l+nk=v_p\left(b^{n-1}\,\prod_{j=0}^{n-1}\,(a+jb)\right)$. 
Now, suppose that $k \geq l$.  Then, it is evident that, for every $s=1,2,\ldots$, the congruence $$a+jb\equiv 0\,\pmod{p^{l+s}}$$ has at least $\left\lfloor\dfrac{n}{p^s}\right\rfloor$ solutions $j\in\{0,1,2,\ldots,n-1\}$.  That is,
$$\sum_{j=0}^{n-1}\,v_p(a+jb)\geq nl+\sum_{s=1}^{\infty}\,\left\lfloor\frac{n}{p^s}\right\rfloor=nl+v_p(n!)\geq v_p(n!)\,.$$
Ergo, we again obtain $\displaystyle v_p(n!)\leq v_p\left(b^{n-1}\,\prod_{j=0}^{n-1}\,(a+jb)\right)$. 
That is, in all cases, $\displaystyle v_p(n!)\leq v_p\left(b^{n-1}\,\prod_{j=0}^{n-1}\,(a+jb)\right)$.  Because $p$ is arbitrary, we conclude that $n!$ divides $\displaystyle b^{n-1}\,\prod_{j=0}^{n-1}\,(a+jb)$.  I believe that there should be a combinatorial proof of this statement, and hope to see it.
