Prove for all $x, y \in \Bbb N$ , $x \mid y (y + 1) \cdots (y + (x - 1))$ I need your help with the following proof:
Prove for all $x, y \in \Bbb N$ 
$$x\mid y  (y + 1) \cdots (y + (x - 1)).$$
What I have so far: 
Proof with induction over $x$.
Let $y$ be arbitrary but fixed.
IB: $x = 1$, then $1 \mid y$.
IH: let $x$ be arbitrary but fixed with $x \mid y (y + 1)\cdots (y + (x - 1))$.
Now we have to prove $(x + 1)\mid  y  (y + 1) \cdots (y + (x - 1))  (y + x)$.
From the IH we know that $x\mid y  (y + 1)\cdots (y + (x - 1))$ and we can conclude that 
$x \mid y (y + 1) \cdots (y + (x - 1))  (y + x)$.
So far I have come, but how do I continue and prove divisibility by $x + 1$?
Any help or hint would be really appreciated.
All the best!
 A: Approach 1:


*

*$x>y$: 


Suppose firstly $x>y$, then $x=y+k, k\in \mathbb{N}$. And of course $k\leq x-1.$


*

*$x<y$: 


Suppose $x<y$ then $x=y-k, k\in \mathbb{N}.$ But you have exactly $x$ consecutive number $y, y+1,..., y+x-1.$ By simple logic we have that one of them have to be $a\cdot x$ for some $a\in \mathbb{N}.$


*

*$x=y$:


Suppose $x=y$, then it's trivial.
Approach 2:
I find that second approach very pleasing. 
Note that you can rewrite: $$y(y+1)\cdot \dots\ \cdot (y+x-1)=\frac{(y+x-1)!}{(y-1)!}.$$ 

And the question is this divisible by $x$?

We can say much more about it:  it's divisible by $x!$. Because:
$$\frac{(y+x-1)!}{(y-1)!x!}=\binom{y+x-1}{x}.$$ And hence it is of course divisible by $x$. 
A: Do the induction on $y$. The case $y=1$ is obvious, as
$$
1(1+1)\dotsm(1+x-1)=x!
$$
is obviously divisible by $x$.
Suppose the case $y$; then
$$
\overbrace{(y+1)\dotsm(y+x-1)}^{\text{common factor}}\,(y+x)-
\underbrace{y\,\overbrace{(y+1)\dotsm(y+x-1)}^{\text{common factor}}}_{\text{divisible by $x$}}=
(y+1)\dotsm(y+x-1)(y+x-y)
$$
