Prove that the product of any $5$ consecutive integers is divisible by 120.

Question

I am going through the Book of Proof, by Richard Hammack, and came across the following exercise: prove that the product of any $5$ consecutive integers is divisible by 120.

My proof was the following:

We prove the more general fact that for $n\in\mathbb{N}$, the product of $n$ consecutive numbers is divisible by $n!$.

Note that the product of $m$ consecutive integers can be written as $n(n-1)(n-2)\ldots(n-m+1)$. Since $\binom{n}{m}$ is by definition an integer, and \begin{align} \binom{n}{m}&=\frac{n!}{m!(n-m)!}\nonumber\\ &=\frac{n(n-1)(n-2)\ldots(n-m+1)(n-m)!}{m!(n-m)!}\nonumber\\ &=\frac{n(n-1)(n-2)\ldots(n-m+1)}{m!}.\nonumber \end{align}

It follows that $m!\mid(n(n-1)(n-2)\ldots(n-m+1))$, where $n(n-1)(n-2)\ldots(n-m+1)$ is the product of $m$ consecutive numbers.

Applying the general fact to $m=5$ proves that The product of any $5$ consecutive integers is divisible by $5!=120$. $\blacksquare$

Is this correct? The solution in the end of the book used a different aproach…