TITLE: Formal proof $\binom{n}{k}$ is an integer
QUESTION [8 upvotes]: In mathematics one defines:
$\left(\begin{array}{c}n\\k\end{array}\right)=\displaystyle\frac{n!}{k!\cdot (n-k)!}$
This is the number of combinations of $k$ elements from a collection of $n$ elements.
I was wondering if it is possible to prove that the result is an integer in a formal way. It's easy to see that $k!$ in the denominator divides the numerator, but how do you prove that the denominator $k!$ also divides the product of all elements between $n-k+1$ and $n$?

REPLY [7 votes]: $$
\frac{n!}{k!\cdot (n-k)!}
=\frac{n(n-1)\cdots(n-k+1)}{k!}
=\frac{(n-k+k)(n-1)\cdots(n-k+1)}{k!}\\
=\frac{(n-1)\cdots(n-k)}{k!}+
\frac{(n-1)\cdots(n-k+1)}{(k-1)!}
$$
Now use (double) induction.
This is the relation that defines Pascal's triangle:
$${n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}.$$