Arithmetic progression - terms divisible by a prime. 
If $p$ is a prime and $p \nmid b$, prove that in the arithmetic progression $a, a+b, a +2b, $ $a+3b, \ldots$, every $p^{th}$ term is divisible by $p$.

I am given the hint that because $\gcd(p,b)=1$, there exist integers $r$ and $s$ satisfying $pr+bs=1$.  Put $n_k$ = $kp-as$ for $k = 1,2,...$ and show that $p \mid a + n_k b$.
I get how to solve the problem once I apply the hint, but I am unsure how to prove the hint.  How do I know what to set $n_k$ to?

hint: define 
$$
f(n) \equiv_p \prod_{j=n}^{n+p-1} (a+bj)
$$
as $(b,p)=1$ we can find an integer $k$ such that $kb \equiv _p 1$. thus (using Fermat's Little Theorem for the second step)
$$
kf(n) \equiv_p k^p f(n) \equiv_p k^p\prod_{j=n}^{n+p-1} (a+bj) \equiv_p \prod_{j=n}^{n+p-1} (ka+j) \equiv_p 0
$$
since the product of any $p$ consecutive integers must be divisible by $p$
