Let $n\in\mathbb{N}$ and let $k\in\{0,\ldots,n\}$. Explain why it follows from
$$\binom nk=\frac{n}{1}\times\frac{n-1}{2}\times\frac{n-2}{3}\times\cdots\times\frac{n-k+1}{k}$$ that $$\binom nk=\frac{n!}{(n-k)!k!}$$
We know $n! = n(n-1)(n-2)...(n-k+1)(n-k)!$. There are $k$ patterns in the product $n(n-1)...(n-k+1)$, namely $n-1+1$, $n-2+1$, $n-3+1$, $\ldots$ , $n-k+1$. A minor point to notice is that in the special case where $k = 0$, then it is a product of no factors, so its value is $1$. If $n - 0$, then $k$ can only be $0$, so again is a product of no factors and its value is $1$.
$$\frac{n}{1}\cdot\frac{n-1}{2}\cdot\frac{n-2}{3}\cdot...\cdot\frac{n-k+1}{k} = \frac{n}{1}\cdot\frac{n-1}{2}\cdot\frac{n-2}{3}\cdot...\cdot\frac{n-k+1}{k}\cdot\frac{(n-k)!}{(n-k)!}=\frac{n!}{k!(n-k)!}$$