TITLE: Prove that $(a+1)(a+2)...(a+b)$ is divisible by $b!$
QUESTION [5 upvotes]: The problem is following, prove that:
$$(a+1)(a+2)...(a+b)\text{ is divisible by } b!\text{ for every positive integer a,b}$$

I've tried solving this problem using mathematical induction, but I don't think that i did it correctly. Here's what i've done
$1.\ b=1\ (Basis)$
$$b!| (a+1)(a+2)...(a+b)$$
$$1 | a+1 \text{, which is true}$$
$2.\ b=k\ (Induction\ Hypothesis)$
$$k! | (a+1)(a+2)...(a+k)\text{, we assume it's true}$$
$$k!*n = (a+1)(a+2)...(a+k)\text{, n is some positive integer}$$
$3.\ b=k+1\ (Inductive\ Step)$
In order to prove I should get:
$$(k+1)!*m = (a+1)(a+2)...(a+k)(a+k+1)\text{, where m is some positive integer}$$
$$(a+1)(a+2)...(a+k)(a+k+1) = k!*n*(a+k+1)$$
$$(a+1)(a+2)...(a+k)(a+k+1) = (k+1)!*n + k!*n*a$$
$$(a+1)(a+2)...(a+k)(a+k+1) - a(a+1)(a+2)...(a+k) = (k+1)!*n$$
$$(a+1)(a+2)...(a+k)(k+1) = (k+1)!*n$$
And as you can see I'm returning at the beginning.
Also I've tried to use the fact that in b consecutive numbers, there must be at least one that divides b, but since i eliminate that number (because I couldn't be sure that the quotient is divisible with (b-1) or (b-2) or ... or 2. And after this i can't continue with (b-1) using this method, because it's not necessary for the other (b-1) to be consecutive.

REPLY [10 votes]: You can do induction simultaneously on $a$ and $b$.
Basis. If $a=0$ the statement is clearly true. If $b=0$ it is also true (an empty product is equal to one).
Inductive hypothesis. Assume that the statement is true if we decrease $a$ or $b$ by one. More precisely, let $L(a,b)$ denote the statement for given $a$ and $b$. We assume that $L(a-1,b)$ and $L(a,b-1)$ are true.
Inductive step. Write:
$$
\begin{align}
        & (a+1) \ldots (a+b-1) (a+b) \\
= \, a  & (a+1) \ldots (a+b-1) \\
+ \quad & (a+1) \ldots (a+b-1) b \\
\end{align}
$$
The first term divides $b!$ because $L(a-1,b)$ is true. The second term divides $b!$ because $L(a,b-1)$ is true.