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I want to use induction to show that the product of any $n$ consecutive positive integers is divisible by $n!$.

I am given the following identity: $$\frac{m(m+1)\cdots(m+n-1)}{n!} = \frac{(m-1)m(m+1)\cdots(m+n-2)}{n!} + \frac{m(m+1)\cdots(m+n-2)}{(n-1)!}$$

The base case is obvious. For the induction hypothesis, assume that for a fixed $k \in \mathbb{Z}^+$, we have: $$\frac{m(m+1)\cdots(m+k-1)}{k!}$$ is an integer.

So, for $n=k+1$, we have: \begin{align*} \frac{m(m+1)\cdots(m+k)}{(k+1)!} &= \frac{(m-1)m(m+1)\cdots(m+k-1)}{(k+1)!} + \frac{m(m+1)\cdots(m+k-1)}{k!}\\ &= \frac{(m-1)m(m+1)\cdots(m+k-1)}{(k+1)!} + p,\ \text{for some}\ p \in \mathbb{Z}^+,\ \text{by the hypothesis.} \end{align*}

Simplifying the right-hand side, we get: $$\frac{(m-1)m(m+1)\cdots(m+k-1)}{(k+1)!} + p = \frac{p(m-1)}{k+1} + p = p\left(\frac{m-1}{k+1 }+1\right) $$

But I got stuck with this expression.

OGreeni
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  • Suggest you beef up your inductive assumption to say that all products of $k-1$ consecutive integers are divisible by $(k-1)!.$ That way you can use both terms on the right of your given identity in the inductive step. – coffeemath Apr 30 '23 at 05:39

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