Prove that for all $n, a \in \mathbb Z$,
$n!|(a+1)(a+2)...(a+n)$.
I was thinking I could do this by induction but I'm a little stuck. Here's what I have:
Basis:
For $n=1$, $a \in \mathbb Z$,
$n! = 1$ and $(a+1)$. $1$ obviously divides any integer so this is true.
Induction Step:
Suppose $k$ such that $k!|(a+1)...(a+k)$ is true, $k,a \in \mathbb Z$.
Then prove for $k+1$. We know that $k!|(k+1)!$ and $k!|(a+1)...(a+k)(a+k+1)$.
But now I'm not sure what to do from here. Or if maybe there's an easier way to prove this..
